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#### Page No 440:

#### Question 1:

If neutrons exert only attractive force, why don't we have a nucleus containing neutrons alone?

#### Answer:

Nuclear forces are short range strong attractive forces that act between two proton-proton, neutron-proton and neutron-neutron pairs. Two protons have strong nuclear forces between them and also exert electrostatic repulsion on each other. However, electrostatic forces are long ranged and have very less effect as compared to the strong nuclear forces.

So, in a nucleus (that is very small in dimension), there's no such significance of repulsive force as compared to the strong attractive nuclear force. On the other hand, an atom contains electrons revolving around its nucleus. These electrons are kept in their orbit by the strong electrostatic force that is exerted on them by the protons present inside the nucleus. Hence, a nucleus contains protons as well as neutrons.

#### Page No 440:

#### Question 2:

Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs?

#### Answer:

Neutrons are chargeless particles and they exert only short range nuclear forces on each other. If we have two pairs of neutrons and the separation between them is same in both the pairs. The force between the neutrons will be of same magnitude for the two pairs until there is some other influence on any of them.

#### Page No 440:

#### Question 3:

A molecule of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why?

#### Answer:

Inside the nucleus, two protons exert nuclear force on each other. These forces are short-ranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (long-ranged). While discussing the behaviour of a hydrogen molecule, the nuclear force between the two protons is always neglected. This is because the separation between the two protons in the molecule is $~$70 pm which is much greater than the range of the nuclear force.

#### Page No 440:

#### Question 4:

Is it easier to take out a nucleon (a) from carbon or from iron (b) from iron or from lead?

#### Answer:

Binding energy per nucleon of a nucleus is defined as the energy required to break-off a nucleon from it.

(a) As the binding energy per nucleon of iron is more than that of carbon, it is easier to take out a nucleon from carbon than iron.

(b) As the binding energy per nucleon of iron is more than that of lead. Therefore, it is easier to take out a nucleon from lead as compared to iron.

#### Page No 440:

#### Question 5:

Suppose we have 12 protons and 12 neutrons. We can assemble them to form either a ^{24}Mg nucleus or two ^{12}C nuclei. In which of the two cases more energy will be liberated?

#### Answer:

If we assemble 6 protons and 6 neutrons to form ^{12}C nucleus, 92.15 MeV (product of mass number and binding energy per nucleon of carbon-12) of energy is released. Therefore, the energy released in the formation of two carbon nuclei is 184.3 MeV. On the other hand, when 12 protons and 12 neutrons are combined to form a ^{24}Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of ^{24}Mg nucleus, more energy is liberated.

#### Page No 440:

#### Question 6:

What is the difference between cathode rays and beta rays? When the two are travelling in space, can you make out which is the cathode ray and which is the beta ray?

#### Answer:

Cathode rays consist of electrons that are accelerated using electrodes. They do not carry high energy and do not harm human body. On the other hand, beta rays consist of highly energetic electrons that can even penetrate and damage human cells. Beta rays are produced by the decay of radioactive nuclei. If the two are travelling in space, they can be distinguished by the phenomenon named production of Bremsstrahlung radiation, which is produced by the deceleration of a high energy particle when deflected by another charged particle, leading to the emission of blue light. Only beta rays are capable of producing it.

#### Page No 440:

#### Question 7:

If the nucleons of a nucleus are separated from each other, the total mass is increased. Where does this mass come from?

#### Answer:

When the nucleons of a nucleus are separated, a certain amount of energy is to be given to the nucleus, which is known as the binding energy.

Binding energy = [(Number of nucleons) $\times $ (Mass of a nucleon) - (Mass of the nucleus)]

When the nucleons of a nucleus are separated, the increase in the total mass comes from the binding energy, which is given to the nucleus to break-off its constituent nucleons as energy is related to mass by the relation given below.

*E = *Δ*mc*^{2}

#### Page No 440:

#### Question 8:

In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged?

#### Answer:

In beta decay, a neutron from the nucleus is converted to a proton releasing an electron and an antineutrino or a proton is converted to a neutron releasing a positron and a neutrino.

$\mathrm{i.e.}{\beta}^{-}\mathrm{decay}:n\to p+e+\overline{\nu}\phantom{\rule{0ex}{0ex}}{\beta}^{+}\mathrm{decay}:p\to n+{e}^{+}+\nu \phantom{\rule{0ex}{0ex}}$

Since the number of valence electrons present in the parent atom do not change, the remaining atom does not get oppositely charged. Instead, due to a change in the atomic number, there's a formation of a new element.

#### Page No 440:

#### Question 9:

When a boron nucleus $\left({}_{5}{}^{10}\mathrm{B}\right)$ is bombarded by a neutron, a α-particle is emitted. Which nucleus will be formed as a result?

#### Answer:

It is given that when a boron nucleus $\left({}_{5}{}^{10}\mathrm{B}\right)$ is bombarded by a neutron, an α-particle is emitted.

Let *X* nucleus be formed as a result of the bombardment.

According to the charge and mass conservation,

${}_{5}{}^{10}B+{}_{0}{}^{1}n\to X+{}_{2}{}^{4}He\phantom{\rule{0ex}{0ex}}\mathrm{Charge}:50\overline{)3}2\phantom{\rule{0ex}{0ex}}\mathrm{Mass}:101\overline{)7}4\phantom{\rule{0ex}{0ex}}$

The mass number of X should be 7 and its atomic number should be 3.

$\therefore X={}_{3}{}^{7}Li$

#### Page No 440:

#### Question 10:

Does a nucleus lose mass when it suffers gamma decay?

#### Answer:

Gamma rays consist of photons that are produced when a nucleus from its excited state comes to its ground state releasing energy. Since gamma rays are chargeless and massless particles, the nucleus does not suffer any loss in mass during the gamma decay.

#### Page No 440:

#### Question 11:

In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Which one has greater liner momentum?

#### Answer:

Two photons having equal liner momentum have equal wavelengths as here for both the photons the direction and magnitude of linear momentum will be same. For the rest of the options, magnitude will be same but nothing can be said about the direction of the photons.

Hence the correct option is D.

#### Page No 440:

#### Question 12:

If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimise the energy?

#### Answer:

When three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.

#### Page No 440:

#### Question 1:

The mass of a neutral carbon atom in ground state is

(a) exact 12 u

(b) less than 12 u

(c) more than 12 u

(d) depends on the form of carbon such as graphite of charcoal.

#### Answer:

(a) exact 12 u

In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.

It is defined such that

1 u = $\frac{1}{12}\times $ (Mass of neutral carbon atom in its ground state)

Mass of neutral carbon atom in its ground state = 12 × 1 u = 12 u

Thus, the mass of neutral carbon atom in its ground state is exactly 12 u.

#### Page No 440:

#### Question 2:

The mass number of a nucleus is equal to

(a) the number of neutrons in the nucleus

(b) the number of protons in the nucleus

(c) the number of nucleons in the nucleus

(d) none of them.

#### Answer:

(c) the number of nucleons in the nucleus

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus.

#### Page No 440:

#### Question 3:

As compared to ^{12}C atom, ^{14}C atom has

(a) two extra protons and two extra electrons

(b) two extra protons but no extra electrons

(c) two extra neutrons and no extra electron

(d) two extra neutrons and two extra electron.

#### Answer:

(c) two extra neutrons and no extra electron

^{12}C and ^{14}C are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons. Therefore, ^{12}C has 6 protons, 6 electrons and 6 neutrons, whereas ^{14}C has 6 electrons, 6 protons and 8 neutrons.

#### Page No 440:

#### Question 4:

The mass number of a nucleus is

(a) always less than its atomic number

(b) always more than its atomic number

(c) equal to its atomic number

(d) sometimes more than and sometimes equal to its atomic number.

#### Answer:

(d) sometimes more than and sometimes equal to its atomic number

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus (like that of hydrogen ^{1}H_{1}), only protons are present. Due to this, the mass number is equal to the atomic number.

#### Page No 440:

#### Question 5:

The graph of ln(*R*/*R*_{0}) versus ln *A*(*R* = radius of a nucleus and *A* = its mass number) is

(a) a straight line

(b) a parabola

(c) an ellipse

(d) none of them.

#### Answer:

(a) a straight line

The average nuclear radius (*R*) and the mass number of the element (*A*) has the following relation:

$R={R}_{o}{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\phantom{\rule{0ex}{0ex}}\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{${R}_{o}$}\right.={A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{${R}_{o}$}\right.\right)=\frac{1}{3}\mathrm{ln}A\phantom{\rule{0ex}{0ex}}$

Therefore, the graph of ln(*R/**R*0) versus ln *A* is a straight line passing through the origin with slope 1/3.

#### Page No 440:

#### Question 6:

Let *F _{pp}*,

*F*and

_{pn}*F*denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. neglect gravitational force. When the separation is 1 fm.

_{nn}(a)

*F*>

_{pp}*F*=

_{pn}*F*

_{nn}(b)

*F*=

_{pp}*F*=

_{pn}*F*

_{nn}(c)

*F*>

_{pp}*F*>

_{pn}*F*

_{nn}(d)

*F*<

_{pp}*F*=

_{pn}*F*

_{nn}#### Answer:

(d) *F*pp < *F*pn = *F*nn

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other. Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of 1 fm.

∴ *F*pp < *F*pn = *F*nn

Here, *F*pp, *F*pn and *F*nn denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

#### Page No 440:

#### Question 7:

Let *F _{pp}*,

*F*and

_{pn}*F*denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. When the separation is 1 fm,

_{nn}(a)

*F*>

_{pp}*F*=

_{pn}*F*

_{nn}(b)

*F*=

_{pp}*F*=

_{pn}*F*

_{nn}(c)

*F*>

_{pp}*F*>

_{pn}*F*

_{nn}(d)

*F*<

_{pp}*F*=

_{pn}*F*

_{nn}#### Answer:

(b) *F*_{pp} = *F*_{pn} = *F*_{nn}

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.

∴ *F*_{pp} = *F*_{pn} = *F*_{nn}

Here, *F*_{pp}, *F*_{pn} and *F*_{nn} denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

#### Page No 440:

#### Question 8:

Two protons are kept at a separation of 10 nm. Let *F _{n}* and

*F*be the nuclear force and the electromagnetic force between them.

_{e}(a)

*F*=

_{e}*F*

_{n}(b)

*F*>>

_{e}*F*

_{n}(c)

*F*<<

_{e}*F*

_{n}(d)

*F*and

_{e}*F*differ only slightly.

_{n}#### Answer:

(b) *F*_{e} >> *F*_{n}

Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10 nm, the electromagnetic force is greater than the nuclear force, i.e.* F*_{e} >> *F*_{n}*. *

#### Page No 441:

#### Question 9:

As the mass number *A* increases, the binding energy per nucleon in a nucleus

(a) increases

(b) decreases

(c) remains the same

(d) varies in a way that depends on the actual value of *A*.

#### Answer:

(d) varies in a way that depends on the actual value of A

Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7 MeV for* A* (50−80). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5 MeV.

#### Page No 441:

#### Question 10:

Which of the following is a wrong description of binding energy of a nucleus?

(a) It is the energy required to break a nucleus into its constituent nucleons.

(b) It is the energy made available when free nucleons combine to form a nucleus.

(c) It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus.

(d) It is the sum of the kinetic energy of all the nucleons in the nucleus.

#### Answer:

(d) It is the sum of the kinetic energies of all the nucleons present in the nucleus.

Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the *Q*-value of the breaking of nucleus, i.e. the difference between the rest energies of reactants (nucleus) and the products (nucleons) or the difference between the kinetic energies of the products and the reactants.

#### Page No 441:

#### Question 11:

In one average-life,

(a) half the active nuclei decay

(b) less than half the active nuclei decay

(c) more than half the active nuclei decay

(d) all the nuclei decay.

#### Answer:

(c) more than half the active nuclei decay

The average life is the mean life time for a nuclei to decay.

It is given as

$\tau =\frac{1}{\lambda}=\frac{{{\rm T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{0.693}$

Here, $\tau ,\lambda \mathrm{and}{{\rm T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

#### Page No 441:

#### Question 12:

In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?

(a) Proton

(b) Neutron

(c) Electron

(d) Photon

#### Answer:

(d) Photon

The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a beta-decay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.

#### Page No 441:

#### Question 13:

During a negative beta decay,

(a) an atomic electron is ejected

(b) an electron which is already present within the nucleus is ejected

(c) a neutron in the nucleus decays emitting an electron

(d) a proton in the nucleus decays emitting an electron.

#### Answer:

(c) a neutron in the nucleus decays emitting an electron

Negative beta decay is given as

$n\to p+{e}^{-}+\overline{\nu}$

Neutron decays to produce proton, electron and anti-neutrino.

#### Page No 441:

#### Question 14:

A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

(a) 6 h

(b) 12 h

(c) 24 h

(d) 128 h.

#### Answer:

(b) 12 h

A freshly prepared radioactive source emits radiation of intensity that is 64 times the permissible level. This means that it is possible to work safely till 6 half-lives (as 2^{6} = 64) of the radioactive source. Since the half-life of the source is 2h, the minimum time after which it would be possible to work safely with this source is 12 h.

#### Page No 441:

#### Question 15:

The decay constant of a radioactive sample is λ. The half-life and the average-life of the sample are respectively

(a) 1/λ and (In 2/λ)

(b) (In 2/λ) and 1/λ

(c) λ(In 2) and 1/λ

(d) λ/(In 2) and 1/λ.

#### Answer:

(b) (ln 2/λ) and 1/λ

The half-life of a radioactive sample (${t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$) is defined as the time elapsed before half the active nuclei decays.

Let the initial number of the active nuclei present in the sample be *N*_{0}.

$\frac{{N}_{o}}{2}={N}_{o}{e}^{-\lambda {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\mathrm{ln}2}{\lambda}$

Average life of the nuclei, ${t}_{av}=\frac{S}{{N}_{o}}=\frac{1}{\lambda}$

Here, *S* is the sum of all the lives of all the *N* nuclei that were active at *t* = 0 and $\lambda $ is the decay constant of the sample.

#### Page No 441:

#### Question 16:

An α-particle is bombarded on ^{14}N. As a result, a ^{17}O nucleus is formed and a particle is emitted. This particle is a

(a) neutron

(b) proton

(c) electron

(d) positron.

#### Answer:

(b) proton

If an alpha particle is bombarded on a nitrogen (N-14) nucleus, an oxygen (O-17) nucleus and a proton are released.

According to the conservation of mass and charge,

${}_{2}{}^{4}He+{}_{7}{}^{14}N\to {}_{6}{}^{17}O+{}_{1}{}^{1}p$

So, the emitted particle is a proton.

#### Page No 441:

#### Question 17:

Ten grams of ^{57}Co kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly

(a) 10 g

(b) 5 g

(c) 2.5 g

(d) 1.25 g.

#### Answer:

(a) 10 g

^{57}Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11$\times $10^{-31 }kg) as compared to the Co atom. Therefore, after 570 days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly 10 g.

#### Page No 441:

#### Question 18:

Free ^{238}U nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes *x* in time *t* after the decay. If a decay takes place when the train is moving at a uniform speed *v*, the distance between the alpha particle and the recoiling nucleus at a time *t* after the decay, as measured by the passenger will be

(a) *x* + *vt*

(b) *x* − *vt*

(c) *x*

(d) depends on the direction of the train.

#### Answer:

(c) *x*

When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is *x* in time *t* after the decay. Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, the separation between the alpha particle and nucleus will be *x*. This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.

#### Page No 441:

#### Question 19:

During a nuclear fission reaction,

(a) a heavy nucleus breaks into two fragments by itself a light nucleus bombarded by thermal neutrons breaks up

(b) a light nucleus bombarded by thermal neutrons breaks up

(c) a heavy nucleus bombarded by thermal neutrons breaks up

(d) two light nuclei combine to give a heavier nucleus and possible other products.

#### Answer:

(c) a heavy nucleus bombarded by thermal neutrons breaks up

In a nuclear reactor, a large fissile atomic nucleus like uranium-235 absorbs a thermal neutron and undergoes a nuclear fission reaction. The heavy nucleus splits into two or more lighter nuclei releasing gamma radiation and free neutrons.

#### Page No 441:

#### Question 1:

As the mass number *A* increases, which of the following quantities related to a nucleus do not change?

(a) Mass

(b) Volume

(c) Density

(d) Binding energy

#### Answer:

(c) Density

Radius of a nucleus with mass number *A* is given as

$R={R}_{o}{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Here, *R _{o}* = 1.2 fm

∴ Volume of the nucleus = $\frac{4\pi {R}^{3}}{3}=\frac{4\pi {{R}_{o}}^{3}A}{3}$

This depends on

*A.*With an increase in A,

*V*increases proportionally.

Mass of the nucleus $\simeq $

*Am*

_{N}Here,

*m*is the mass of a nucleon.

_{N}_{ }Therefore, mass of the nucleus also increases with the increasing mass number.

Binding energy also depends on mass number (number of nucleons) as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.

On the other hand,

Density = $\frac{\mathrm{Mass}}{\mathrm{Volume}}=\frac{A{m}_{N}}{{\displaystyle \raisebox{1ex}{$4\pi {R}^{3}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{A{m}_{N}}{{\displaystyle \raisebox{1ex}{$4\pi {{R}_{o}}^{3}A$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{{m}_{N}}{{\displaystyle \raisebox{1ex}{$4\pi {{R}_{o}}^{3}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{3{m}_{N}}{{\displaystyle 4\pi {{R}_{o}}^{3}}}$

This is independent of

*A*and hence does not change as mass number increases.

#### Page No 441:

#### Question 2:

The heavier nuclei tend to have larger *N/Z* ratio because

(a) a neutron is heavier than a proton

(b) a neutron is an unstable particle

(c) a neutron does not exert electric repulsion

(d) Coulomb forces have longer range compared to the nuclear forces.

#### Answer:

(c) a neutron does not exert electric repulsion

(d) Coulomb forces have longer range compared to the nuclear forces

This is because in heavy nuclei, the *N/Z* ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons.The neutrons exert only attractive short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio (*N/Z*).

#### Page No 441:

#### Question 3:

A free neutron decays to a proton but a free proton does not decay to a neutron. This is because

(a) neutron is a composite particle made of a proton and an electron whereas proton is a fundamental particle

(b) neutron is an uncharged particle whereas proton is a charged particle

(c) neutron has large rest mass than the proton

(d) weak forces can operate in a neutron but not in a proton.

#### Answer:

(c) neutron has large rest mass than the proton.

A nucleus is made up of two fundamental particles- neutrons and protons. If a nucleus has more number of neutrons than what is needed to have stability, then neutrons decay into protons and if there's an excess of protons, then they decay to form neutrons. Since a neutron has larger rest mass than a proton, the *Q*-value of its decay reaction is positive and a free neutron decays to a proton, while an isolated proton cannot decay to a neutron as the *Q*-value of its decay reaction is negative. Hence, it is physically not possible.

#### Page No 441:

#### Question 4:

Consider a sample of a pure beta-active material.

(a) All the beta particles emitted have the same energy.

(b) The beta particles originally exist inside the nucleus and are ejected at the time of beta decay.

(c) The antineutrino emitted in a beta decay has zero mass and hence zero momentum.

(d) The active nucleus changes to one of its isobars after the beta decay.

#### Answer:

(d) The active nucleus changes to one of its isobars after the beta decay.

In a beta decay, either a neutron is converted to a proton or a proton is converted to a neutron such that the mass number does not change. Also,the number of the nucleons present in the nucleus remains the same. Thus, the active nucleus gets converted to one of its isobars after beta decay.

#### Page No 441:

#### Question 5:

In which of the following decays the element does not change?

(a) α-decay

(b) β^{+}-decay

(c) β^{−}-decay

(d) γ-decay

#### Answer:

(d) γ-decay

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number *Z* by 4 and neutron number *N* by 2 such that the element gets changed.

${}_{Z}{}^{A}X\to {}_{Z-2}{}^{A-4}Y+{}_{2}{}^{4}He$

During β^{−}-decay, a neutron is converted to a proton, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.

${}_{Z}{}^{A}X\to {}_{Z+1}{}^{A}Y+e+\overline{\nu}$

During β^{+}-decay, a proton in the nucleus is converted to a neutron, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.

${}_{Z}{}^{A}X\to {}_{Z-1}{}^{A}Y+{\beta}^{+}+\nu $

When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn't change.

#### Page No 441:

#### Question 6:

In which of the following decays the atomic number decreases?

(a) α-decay

(b) β^{+}-decay

(c) β^{−}-decay

(d) γ-decay

#### Answer:

(a) α-decay

(b) β^{+}-decay

In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) *Z* as well as neutron number *N* by 2.

${}_{Z}{}^{A}X\to {}_{Z-2}{}^{A-4}Y+{}_{2}{}^{4}He$

During β^{−}-decay, a neutron is converted to a proton, an electron and an antineutrino. Thus, there is an increase in the atomic number.

${}_{Z}{}^{A}X\to {}_{Z+1}{}^{A}Y+{e}^{-}+\overline{\nu}$

During β^{+}-decay, a proton in the nucleus is converted to a neutron, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number.

${}_{Z}{}^{A}X\to {}_{Z-1}{}^{A}Y+{\beta}^{+}+\nu $

When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn't change.

Therefore, alpha and beta plus decay suffer decrease in atomic number.

#### Page No 441:

#### Question 7:

Magnetic field does not cause deflection in

(a) α-rays

(b) beta-plus rays

(c) beta-minus rays

(d) gamma rays

#### Answer:

(d) gamma rays

Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as *$\left|\overrightarrow{F}\right|=\left|q(\overrightarrow{v}\times \overrightarrow{B})\right|$. *

Here, *q *is the charge on the particle that is moving with speed *v* in a uniform magnetic field *B*.

Since alpha, beta-plus and beta-minus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.

#### Page No 441:

#### Question 8:

Which of the following are electromagnetic waves?

(a) α-rays

(b) Beta-plus rays

(c) Beta-minus rays

(d) Gamma rays

#### Answer:

(d) Gamma rays

Alpha rays, beta-plus and beta-minus rays carry charged particles that show particle behaviour. On the other hand, gamma rays carry photons that show particle as well as wave behaviour. Hence, only gamma rays are electromagnetic waves.

#### Page No 441:

#### Question 9:

Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because

(a) a lithium nucleus is more tightly bound than a carbon nucleus

(b) carbon nucleus is an unstable particle

(c) it is not energetically favourable

(d) Coulomb repulsion does not allow the nuclei to come very close.

#### Answer:

(d) Coulomb repulsion does not allow the nuclei to come very close.

Lithium atom contains 3 protons and 3 neutrons in the nucleus and 3 valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being short-range are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.

#### Page No 442:

#### Question 10:

For nuclei with *A* > 100,

(a) the binding energy of the nucleus decreases on an average as *A* increases

(b) the binding energy per nucleon decreases on an average as A increases

(c) if the nucleus breaks into two roughly equal parts, energy is released

(d) if two nuclei fuse to form a bigger nucleus, energy is released.

#### Answer:

(b) the binding energy per nucleon decreases on an average as A increases

(c) if the nucleus breaks into two roughly equal parts, energy is released

Binding energy per nucleon varies in a way that it depends on the actual value of mass number (*A)*. As the mass number (*A)* increases, the binding energy also increases and reaches its maximum value of 8.7 MeV for *A* (50−80) and for *A* > 100. The binding energy per nucleon decreases as *A* increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50−80.

#### Page No 442:

#### Question 1:

Assume that the mass of a nucleus is approximately given by *M* = *Am _{p}* where A is the mass number. Estimate the density of matter in kgm

^{−3}inside a nucleus. What is the specific gravity of nuclear matter?

#### Answer:

Given:

Mass of the nucleus, *M *= A*m _{p}*

Volume of the nucleus,

*V*= $\frac{4}{3}\pi {{R}_{0}}^{3}A$

Density of the matter, $d=\frac{M}{V}=\frac{A{m}_{p}}{{\displaystyle \frac{4}{3}}\pi {{R}_{0}}^{3}A}$

$=\frac{3{m}_{p}}{4\times \pi {{R}_{0}}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{3\times 1.007276}{4\times 3.14(1.1{)}^{3}}\phantom{\rule{0ex}{0ex}}=3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^{3}$

Specific gravity of the nuclear matter = $\frac{\mathrm{Density}\mathrm{of}\mathrm{matter}}{\mathrm{Density}\mathrm{of}\mathrm{water}}$

$\therefore $ Specific gravity = $\frac{3\times {10}^{17}}{{10}^{3}}$ = 3 $\times $ 10

^{14}kg/m

^{3}

#### Page No 442:

#### Question 2:

A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0 × 10^{30} kg (twice the mass of the sun).

#### Answer:

Given:

Mass of the neutron star, *M* = 4.0 × 10^{30} kg

Density of nucleus, *d = *2.4$\times $10^{17}${}^{}$

Density of nucleus, $d=\frac{M}{V}$

Here, *V* is the volume of the nucleus.

$\therefore V=\frac{M}{d}=\frac{4\times {10}^{30}}{2.4\times {10}^{17}}\phantom{\rule{0ex}{0ex}}=\frac{1}{0.6}\times {10}^{13}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times {10}^{14}\phantom{\rule{0ex}{0ex}}$

If *R* is the radius, then the volume of the neutron star is given by

$V=\frac{4}{3}\pi {R}^{3}\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{6}\times {10}^{14}=\frac{4}{3}\times \mathrm{\pi}\times {R}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}^{3}=\frac{1}{6}\times \frac{3}{4}\times \frac{1}{\mathrm{\pi}}\times {10}^{14}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}^{3}=\frac{1}{8}\times \frac{100}{\mathrm{\pi}}\times {10}^{12}\phantom{\rule{0ex}{0ex}}\therefore R=\frac{1}{2}\times {10}^{4}\times 3.17\phantom{\rule{0ex}{0ex}}=1.585\times {10}^{4}=15\mathrm{km}$

#### Page No 442:

#### Question 3:

Calculate the mass of an α-particle. Its Its binding energy is 28.2 MeV.

#### Answer:

Given:

Binding energy of α particle = 28.2 MeV

Let *x* be the mass of α particle.

We know an α particle consists of 2 protons and 2 neutrons.

Binding energy, $B=\left(Z{m}_{p}+N{m}_{n}-M\right){c}^{2}\phantom{\rule{0ex}{0ex}}$

Here, *m*_{p} = Mass of proton

*m*_{n} = Mass of neutron

*Z*= Number of protons

*N* = Number of neutrons

*c* = Speed of light

On substituting the respective values, we have

$28.2=(2\times 1.007276+2\times 1.008665-x){c}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=4.0016\mathrm{u}$

#### Page No 442:

#### Question 4:

How much energy is released in the following reaction:

^{7}Li + p → α + α.

Atomic mass of ^{7}Li = 7.0160 u and that of ^{4}He = 4.0026 u.

#### Answer:

Given:

Mass of ^{7}Li = 7.0160 u

Mass of ^{4}He = 4.0026 u.

Reaction:

$\mathrm{L}{i}^{7}+p\mathit{\to}\alpha \mathit{+}\alpha \mathit{+}E\mathit{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Energy release $\left(E\right)$ is given by

$E=\left[m\left({}^{7}\mathrm{Li}\right)+\left({m}_{p}\right)-2\times m\left({}^{4}\mathrm{He}\right)\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left[\left(7.0160\mathrm{u}+1.007276\mathrm{u}\right)-2\left(4.0026\mathrm{u}\right)\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=(8.023273\mathrm{u}-8.0052\mathrm{u}){\mathrm{c}}^{2}\phantom{\rule{0ex}{0ex}}=0.018076\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=16.83\mathrm{MeV}$

#### Page No 442:

#### Question 5:

Find the binding energy per nucleon of ${}_{79}{}^{197}$Au if its atomic mass is 196.96 u.

#### Answer:

Given:

Atomic mass of Au,* A* = 196.96

Atomic number of Au, *Z* = 79

Number of neutrons,* N* = 118

Binding energy, $B=(\mathrm{Z}{m}_{p}+N{m}_{n}-M){c}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here, *m*_{p} = Mass of proton

*M* = Mass of nucleus

*m*_{n} = Mass of neutron

*c* = Speed of light

On substituting the respective values, we get

$B=\left[\right(79\times 1.007276+118\times 1.008665)\mathrm{u}-196.96\mathrm{u}]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left(198.597274-196.96\right)\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=1524.302094\mathrm{MeV}$

Binding energy per nucleon $=\frac{1524.3}{197}=7.737\mathrm{MeV}$

#### Page No 442:

#### Question 6:

(a) Calculate the energy released if ^{238}U emits an α-particle. (b) Calculate the energy to be supplied to ^{238}U it two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U, ^{234}Th and ^{4}He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

#### Answer:

(a)

Given:

Atomic mass of ^{238}U, *m*(^{238}U) = 238.0508 u

Atomic mass of ^{234}Th, *m*(^{234}Th) = 234.04363 u

Atomic mass of ^{4}He, *m*(^{4}He) = 4.00260 u

When ^{238}U emits an α-particle, the reaction is given by

${\mathrm{U}}^{238}\to {\mathrm{Th}}^{234}+{\mathrm{He}}^{4}$

Mass defect, Δ*m* = [*m*(^{238}U)$-$(*m*(^{234}Th)+*m*(^{4}He))]

Δ*m** = *[238.0508 $-$ (234.04363 + 4.00260) = 0.00457 u

Energy released (*E*) when ^{238}U emits an α-particle is given by

$E=\u2206m{c}^{2}\phantom{\rule{0ex}{0ex}}E=[0.00457\mathrm{u}]\times 931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}\Rightarrow E=4.25467\mathrm{MeV}=4.255\mathrm{MeV}\phantom{\rule{0ex}{0ex}}$

(b)

When two protons and two neutrons are emitted one by one, the reaction will be

${\mathrm{U}}^{233}\to {\mathrm{Th}}^{234}+2n+2p$

$\mathrm{Mass}\mathrm{defect},\u2206m=m\left({\mathrm{U}}^{238}\right)-[m\left({\mathrm{Th}}^{234}\right)+2\left({m}_{\mathrm{n}}\right)+2\left({m}_{p}\right)]\phantom{\rule{0ex}{0ex}}\u2206m=238.0508\mathrm{u}-[234.04363\mathrm{u}+2(1.008665)\mathrm{u}+2(1.007276)\mathrm{u}]\phantom{\rule{0ex}{0ex}}\u2206m=0.024712\mathrm{u}$

Energy released (*E*) when ^{238}U emits two protons and two neutrons is given by

$E=\u2206m{c}^{2}\phantom{\rule{0ex}{0ex}}E=0.024712\times 931.5\mathrm{MeV}\phantom{\rule{0ex}{0ex}}E=23.019=23.02\mathrm{MeV}$

#### Page No 442:

#### Question 7:

Find the energy liberated in the reaction

^{223}Ra → ^{209}Pb + ^{14}C.

The atomic masses needed are as follows.

^{223}Ra ^{209}Pb ^{14}C

22..018 u 208.981 u 14.003 u

#### Answer:

Given:

Atomic mass of ^{223}Ra, *m*(^{223}Ra) = 223.018 u

Atomic mass of ^{209}Pb, *m*(^{209}Pb) = 208.981 u

Atomic mass of ^{14}C, *m*(^{14}C) = 14.003 u

Reaction:

${}^{223}\mathrm{Ra}{\to}^{209}\mathrm{Pb}+{}^{14}\mathrm{C}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Energy},E=\left[m\left({}^{223}Ra\right)-\left(m\left({}^{209}Pb\right)+m\left({}^{14}\mathrm{C}\right)\right)\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left[223.018\mathrm{u}-\left(208.981+14.003\right)\mathrm{u}\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=0.034\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=31.65\mathrm{MeV}$

#### Page No 442:

#### Question 8:

Show that the minimum energy needed to separate a proton from a nucleus with *Z* protons and *N* neutrons is

$\u2206E=({M}_{Z-1,N}+{M}_{B}-{M}_{Z,N}){c}^{2}$

where *M _{Z}_{,N}* = mass of an atom with

*Z*protons and

*N*neutrons in the nucleus and

*M*= mass of a hydrogen atom. This energy is known as

_{B}*proton-separation energy*.

#### Answer:

Given:

Mass of an atom with *Z* protons and *N* neutrons = *M _{Z}*,

_{N}Mass of hydrogen atom =

*M*

_{H}

As hydrogen contains only protons, the reaction will be given by

${E}_{Z\mathit{,}N}\to {E}_{\mathrm{Z}-1,N}+{p}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{Z\mathit{,}N}\to {E}_{z-1,N}{+}^{1}{\mathrm{H}}_{1}$

∴ Minimum energy needed to separate a proton, $\u2206E=({M}_{\mathrm{Z}-1,\mathrm{N}}+{M}_{\mathrm{H}}-{M}_{\mathrm{Z},\mathrm{N}}){c}^{2}$

#### Page No 442:

#### Question 9:

Calculate the minimum energy needed to separate a neutron from a nucleus with *Z* protons and *N* neutrons it terms of the masses M_{Z.N}' M_{Z}_{,N−1} and the mass of the neutron.

#### Answer:

Before the separation of neutron, let the mass of the nucleus be *M*_{Z}_{,N}.

Then, after the separation of neutron, the mass of the nucleus will be *M _{Z}*,

_{N-1.}

The reaction is given by

${E}_{Z\mathit{,}N}={E}_{Z,N-1}+{}_{0}{}^{1}\mathrm{n}$

If ${M}_{\mathrm{N}}$ is the mass of the neutron, then the energy needed to separate the neutron b$\left(\u2206E\right)$ will be

$\u2206E$ = (Final mass of nucleus + Mass of neutron − Initial mass of the nucleus)

*c*

^{2}

$\u2206E=({M}_{Z,\mathrm{N}-1}+{M}_{\mathrm{N}}-{M}_{Z,\mathrm{N}}){c}^{2}$

#### Page No 442:

#### Question 10:

^{32}P beta-decays to ^{32}S. Find the sum of the energy of the antineutrino and the kinetic energy of the β-particle. Neglect the recoil of the daughter nucleus. Atomic mass of ^{32}P = 31.974 u and that of ^{32}S = 31.972 u.

#### Answer:

Given:

Atomic mass of ^{32}P,* m*(^{32}P) = 31.974 u

Atomic mass of ^{32}S, m(^{32}S) = 31.972 u

Reaction:

${\mathrm{P}}^{32}\to {\mathrm{S}}^{32}{+}_{1}{\mathrm{v}}^{0}{+}_{-1}{\mathrm{\beta}}^{0}$

Energy of antineutrino and β-particle, *E* = [*m*(^{32}P)$-$*m*(^{32}S)]*c*^{2}

= (31.974 u− 31.972 u)*c*^{2}

= 0.002 × 931 = 1.862 MeV

#### Page No 442:

#### Question 11:

A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.

#### Answer:

Given:

Half-life period of free neutron beta-decays to a proton, ${T}_{1/2}$ = 14 minutes

Half-life period, *T*_{1}_{/2} = $\frac{0.6931}{\lambda}$

Here, $\lambda $ = Decay constant

$\therefore \lambda =\frac{0.693}{14\times 60}\phantom{\rule{0ex}{0ex}}=8.25\times {10}^{-4}{\mathrm{s}}^{-1}$

If *m*_{p} is the mass of proton, let *m*_{n} and *m*_{e} be the mass of neutron and mass of electron, respectively.

$\therefore \; Energ\mathrm{y}\mathrm{liberated},E=[{m}_{n}-\left({m}_{p}+{m}_{e}\right)]{c}^{2}\phantom{\rule{0ex}{0ex}}=[1.008665\mathrm{u}-\left(1.007276+0.0005486\right)\mathrm{u}]{c}^{2}\phantom{\rule{0ex}{0ex}}=0.0008404\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=782\mathrm{keV}\phantom{\rule{0ex}{0ex}}$

#### Page No 442:

#### Question 12:

Complete the following decay schemes.

(a) ${}_{88}{}^{226}\mathrm{Ra}\to \alpha +$

(b) ${}_{8}{}^{19}\mathrm{O}\to {}_{9}{}^{19}\mathrm{F}+$

(c) ${}_{13}{}^{25}\mathrm{Al}\to {}_{12}{}^{25}\mathrm{Mg}+$

#### Answer:

$\left(\mathrm{a}\right){}_{88}{}^{226}\mathrm{Ra}\to {}_{2}{}^{4}\mathrm{\alpha}+{}_{86}{}^{222}\mathrm{Rn}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right){}_{8}{}^{19}\mathrm{O}\to {}_{9}{}^{19}\mathrm{F}+\overline{e}+\overline{v}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right){}_{13}{}^{25}\mathrm{Ar}\to {}_{12}{}^{25}\mathrm{Mg}+{e}^{+}+v$

#### Page No 442:

#### Question 13:

In the decay ^{64}Cu → ^{64}Ni + e^{+} + *v*, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.

(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?

(b) What is the momentum of this neutrino in kg m s^{−1}?

Use the formula applicable to a photon.

#### Answer:

Given:

Maximum kinetic energy of the positron, *K* = 0.650 MeV

(a) Neutrino and positron are emitted simultaneously.

∴ Energy of neutrino = 0.650 − Kinetic energy of the given positron

= 0.650 − 0.150

= 0.5 MeV = 500 keV

(b) Momentum of the neutrino, $P=\frac{E}{c}$

Here, *E* = Energy of neutrino

*c* = Speed of light

$\Rightarrow P=\frac{500\times 1.6\times {10}^{-19}}{3\times {10}^{8}}\times {10}^{3}\phantom{\rule{0ex}{0ex}}=2.67\times {10}^{-22}{\mathrm{Kgms}}^{-1}$

#### Page No 442:

#### Question 14:

Potassium-40 can decay in three modes. It can decay by β^{−}-emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the *Q*-values in each of the three cases. Atomic masses of ${}_{18}{}^{40}\mathrm{Ar},{}_{19}{}^{40}\mathrm{K}\mathrm{and}{}_{20}{}^{40}\mathrm{Ca}$ are 39.9624 u, 39.9640 u and 39.9626 u respectively.

#### Answer:

(a) Decay of potassium-40 by β^{−}emission is given by

${}_{19}\mathrm{K}^{40}{\to}_{20}{\mathrm{Ca}}^{40}+{\mathrm{\beta}}^{-}+\overline{\mathrm{v}}$

Decay of potassium-40 by β^{+} emission is given by

${}_{19}\mathrm{K}^{40}{\to}_{18}{\mathrm{Ar}}^{40}+{\mathrm{\beta}}^{+}+\mathrm{v}$

Decay of potassium-40 by electron capture is given by

${}_{19}\mathrm{K}^{40}+{\mathrm{e}}^{-}{\to}_{18}{\mathrm{Ar}}^{40}+\mathrm{v}$

(b)

*Q*_{value} in the β^{−} decay is given by

*Q*_{value} = [*m*(_{19}K^{40}) − *m*(_{20}Ca^{40})]*c*^{2}

= [39.9640 u − 39.9626 u]*c*^{2}

= 0.0014 $\times $ 931 MeV

= 1.3034 MeV

*Q*_{value} in the β^{+} decay is given by

*Q*_{value} = [*m*(_{19}K^{40}) − *m*(_{20}Ar^{40}) − 2*m*_{e}]*c*^{2}

= [39.9640* u* − 39.9624* u* − 0.0021944 u]*c*^{2}

= (39.9640 − 39.9624) 931 MeV − 1022 keV

= 1489.96 keV − 1022 keV

= 0.4679 MeV

*Q*_{value} in the electron capture is given by

*Q*_{value} = [ *m*(_{19}K^{40}) − *m*(_{20}Ar^{40})]*c*^{2}

= (39.9640 − 39.9624)*uc*^{2}

= 1.4890 = 1.49 MeV

#### Page No 442:

#### Question 15:

Lithium (*Z* = 3) has two stable isotopes ^{6}Li and ^{7}Li. When neutrons are bombarded on lithium sample, electrons and α-particles are ejected. Write down the nuclear process taking place.

#### Answer:

The nuclear process taking place is shown below.

${}_{8}{}^{6}\mathrm{Li}+n\to {}_{3}{}^{7}\mathrm{Li}\phantom{\rule{0ex}{0ex}}{}_{3}{}^{7}\mathrm{Li}+n\to {}_{3}{}^{8}\mathrm{Li}\to {}_{4}{}^{8}\mathrm{Be}+\overline{v}+{e}^{-}\phantom{\rule{0ex}{0ex}}{}_{4}{}^{8}\mathrm{Be}\to {}_{2}{}^{4}\mathrm{He}+{}_{2}{}^{4}\mathrm{He}$

#### Page No 442:

#### Question 16:

The masses of ^{11}C and ^{11}B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of ^{11}C to ^{11}B.

#### Answer:

Given:

Mass of ^{11}C, *m*(^{11}C) = 11.0114 u

Mass of ^{11}B, *m*(^{11}B) = 11.0093 u

Energy liberated in the β^{+} decay (*Q)* is given by

$Q=\left[m\left({}^{11}C\right)-m\left({}^{11}B\right)-2{m}_{e}\right]{c}^{2}$

= (11.0114 u − 11.0093 u $-$ 2$\times $0.0005486 u)*c*^{2}

= 0.0010028 $\times $ 931 MeV

= 0.9336 MeV = 933.6 keV

For maximum KE of the positron, energy of neutrino can be taken as zero.

∴ Maximum KE of the positron = 933.6 keV

#### Page No 442:

#### Question 17:

^{228}Th emits an alpha particle to reduce to ^{224}Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:

${}^{228}\mathrm{Th}\to {}^{224}\mathrm{Ra}*+\alpha \phantom{\rule{0ex}{0ex}}{}^{224}\mathrm{Ra}*{\to}^{224}\mathrm{Ra}+{\rm Y}(217\mathrm{keV}).$

Atomic mass of ^{228}Th is 228.028726 u, that of ^{224}Ra is 224.020196 u and that of ${}_{2}{}^{4}\mathrm{H}$ is 4.00260 u.

#### Answer:

Given:

Atomic mass of ^{228}Th, *m*(^{228}Th) = 228.028726 u

Atomic mass of ^{224}Ra, *m*(^{224}Ra) = 224.020196 u

Atomic mass of ${}_{2}{}^{4}\mathrm{H}$, *m*(${}_{2}{}^{4}\mathrm{H}$) = 4.00260 u

Mass of ^{224}Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, *K *= $\left[m\left({}^{228}\mathrm{Th}\right)-\left[m\left({}^{224}\mathrm{Ra}\right)+m\left({}_{2}{}^{4}\mathrm{H}\right)\right]\right]$c^{2}

= (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]

= 5.30383 MeV = 5.304 MeV

#### Page No 442:

#### Question 18:

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:

^{12}N → ^{12}C* + *e*^{+} + *v*

^{12}C* → ^{12}C + γ (4.43MeV).

The atomic mass of ^{12}N is 12.018613 u.

#### Answer:

Given:

Atomic mass of ^{12}N,* m*(^{12}N) = 12.018613 u

^{12}N → ^{12}C* + *e*^{+} + *v*

^{12}C* → ^{12}C + γ (4.43 MeV)

Net reaction is given by

^{ }^{12}N → ^{12}C* + e ^{+} + v + *γ

*(4.43 MeV)*

*Q*

_{value}of the ${\beta}^{+}$ decay will be

*Q*

_{value}= [

*m*(

^{12}N) $-$ (

*m*(

^{12}C*) + 2

*m*

_{e})]

*c*

^{2}

= [12.018613 $\times $931 MeV $-$ (12$\times $931 + 4.43) MeV $-$ (2$\times $511) keV]

= [11189.3287 $-$ 11176.43 $-$ 1.022] MeV

= 11.8767 MeV = 11.88 MeV

The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.

#### Page No 442:

#### Question 19:

The decay constant of ${}_{80}{}^{197}\mathrm{Hg}$ (electron capture to ${}_{79}{}^{197}\mathrm{Au}$) is 1.8 × 10^{−4} S^{−1}. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

#### Answer:

Given:

Decay constant of ${}_{80}{}^{197}\mathrm{Hg}$, $\lambda $ = 1.8 × 10^{−4} s^{$-1$}

(a)

Half-life, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda}$

$\Rightarrow {T}_{1/2}=\frac{0.693}{1.8\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=3850\mathrm{s}=64\mathrm{minutes}\phantom{\rule{0ex}{0ex}}$

(b)

$\mathrm{Average}\mathrm{life},{T}_{av}=\frac{{T}_{1/2}}{0.693}\phantom{\rule{0ex}{0ex}}=\frac{64}{0.693}\phantom{\rule{0ex}{0ex}}=92\mathrm{minutes}\phantom{\rule{0ex}{0ex}}$

(c)

Number of active nuclei of mercury at *t* = 0

= *N*_{0}_{ }

= 100

Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = *N* = 75

Now, $\frac{N}{{N}_{0}}={e}^{-\lambda t}$

Here, *N = *Number of inactive nuclei

*N*_{0} = Number of nuclei at* t *= 0

$\lambda $ = Disintegration constant

On substituting the values, we get

$\frac{75}{100}={e}^{-\lambda t}$

$\Rightarrow 0.75={\mathrm{e}}^{-\lambda x}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{In}0.75=-\lambda t\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{\mathrm{In}0.75}{-0.00018}\phantom{\rule{0ex}{0ex}}=1600\mathrm{s}$

#### Page No 443:

#### Question 20:

The half-life of ^{199}Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of ^{198}Au. (b) What will be the activity after 7 days? Take the atomic weight of ^{198}Au to be 198 g mol^{−1}.

#### Answer:

Given:

Half-life of ^{199}Au, *T*_{1}_{/2}= 2.7 days

Disintegration constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle 1/2}}}$ = $\frac{0.693}{2.7\times 24\times 60\times 60}=2.97\times {10}^{-6}{\mathrm{s}}^{-1}$

Number of atoms left undecayed, *N* = $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}$

Now, activity, ${A}_{0}=\lambda N$

= $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}\times 2.97\times {10}^{-6}$

= 0.244 Ci

(b) After 7 days,

Activity, $A={A}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}$

Here, *A*_{0} = 0.244 Ci

$\therefore A=0.244\times {e}^{-2.97\times {10}^{-6}\times 7\times 3600\times 24}\phantom{\rule{0ex}{0ex}}=0.244\times {e}^{17962.56\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=0.040\mathrm{Ci}$

#### Page No 443:

#### Question 21:

Radioactive ^{131}I has a half-life of 8.0 days. A sample containing ^{131}I has activity 20 µCi at *t* = 0. (a) What is its activity at *t* = 4 days? (b) What is its decay constant at *t* = 4.0 days?

#### Answer:

Given:

Half-life of radioactive ^{131}I, *T*_{1}_{/2} = 8 days

Activity of the sample at *t* = 0,* **A*_{0} = 20 µCi

Time, *t* = 4 days

(a) Disintegration constant, $\lambda =\frac{0.693}{{T}_{1/2}}=0.0866$

Activity (*A*) at *t* = 4 days is given by

$A={A}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}\Rightarrow A=20\times {10}^{-6}\times {e}^{-0.0866\times 4}\phantom{\rule{0ex}{0ex}}=1.41\times {10}^{-6}\mathrm{Ci}=14\mathrm{\mu Ci}$

(b) Decay constant is a constant for a radioactive sample and depends only on its half-life.

$\lambda =\frac{0.693}{[8\times 24\times 3600]}\phantom{\rule{0ex}{0ex}}=1.0026\times {10}^{-6}{\mathrm{s}}^{-1}$

#### Page No 443:

#### Question 22:

The decay constant of ^{238}U is 4.9 × 10^{−18} S^{−1}. (a) What is the average-life of ^{238}U? (b) What is the half-life of ^{238}U? (c) By what factor does the activity of a ^{238}U sample decrease in 9 × 10^{9} years?

#### Answer:

Given:

Decay constant, $\lambda $ = 4.9 × 10^{−18} s^{−1}

(a) Average life of uranium $\left(\tau \right)$ is given by

$\tau =\frac{1}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{1}{4.9\times {10}^{-18}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4.9}\times {10}^{18}\mathrm{s}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\mathrm{years}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\mathrm{years}\phantom{\rule{0ex}{0ex}}=6.47\times {10}^{-7}\times {10}^{16}\mathrm{years}\phantom{\rule{0ex}{0ex}}=6.47\times {10}^{9}\mathrm{years}$

(b) Half-life of uranium $\left({T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$ is given by

${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda}=\frac{0.693}{4.9\times {10}^{-18}}\phantom{\rule{0ex}{0ex}}=\frac{0.693}{4.9}\times {10}^{18}\mathrm{s}\phantom{\rule{0ex}{0ex}}=0.1414\times {10}^{18}\mathrm{s}\phantom{\rule{0ex}{0ex}}=\frac{0.1414\times {10}^{18}}{365\times 24\times 3600}\phantom{\rule{0ex}{0ex}}=\frac{1414\times {10}^{12}}{365\times 24\times 36}\phantom{\rule{0ex}{0ex}}=4.48\times {10}^{-3}\times {10}^{12}\phantom{\rule{0ex}{0ex}}=4.5\times {10}^{9}\mathrm{years}\phantom{\rule{0ex}{0ex}}$

(c) Time, *t* = 9 × 10^{9} years

Activity (*A*) of the sample, at any time* t,* is given by

$A=\frac{{A}_{0}}{{\displaystyle {2}^{\frac{t}{{T}_{1/2}}}}}\phantom{\rule{0ex}{0ex}}$

Here, *A*_{0} = Activity of the sample at *t* = 0

$\therefore \frac{{A}_{0}}{A}={2}^{\frac{9\times {10}^{9}}{4.5\times {10}^{9}}}={2}^{2}=4$

#### Page No 443:

#### Question 23:

A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes. (a) What is the decay constant of the sample? (b) What is its half-life?

#### Answer:

Given:

Initial rate of decay, *A*_{0} = 500,

Rate of decay after 50 minutes, *A* = 200

Time, *t* = 50 min

= 50 $\times $ 60

= 3000 s

(a)

Activity, *A *= *A*_{0}*e*^{−λ}^{t}

Here, $\lambda $ = Disintegration constant

$\therefore $ 200 = 500 × *e*^{−50}^{×60×λ}

$\Rightarrow \frac{2}{5}={e}^{-3000\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{In}\frac{2}{5}=-3000\lambda \phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =3.05\times {10}^{-4}{\mathrm{s}}^{-1}\phantom{\rule{0ex}{0ex}}$

(b)

$\mathrm{Half}-\mathrm{life},{T}_{1/2}=\frac{0.693}{\lambda}\phantom{\rule{0ex}{0ex}}=2272.13\mathrm{s}\phantom{\rule{0ex}{0ex}}=38\mathrm{min}$

#### Page No 443:

#### Question 24:

The count rate from a radioactive sample falls from 4.0 × 10^{6} per second to 1.0 × 10^{6} per second in 20 hours. What will be the count rate 100 hours after the beginning?

#### Answer:

Given:

Initial count rate of radioactive sample, *A*_{0} = 4 × 10^{6} disintegration/sec

Count rate of radioactive sample after 20 hours, *A* = 1 × 10^{6} disintegration/sec

Time, *t* = 20 hours

Activity of radioactive sample $\left(A\right)$ is given by

$A=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},{T}_{1/2}=\mathrm{Half}-\mathrm{life}\mathrm{period}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}{A}_{0}\mathrm{and}A,\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}{2}^{\frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}={2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2\phantom{\rule{0ex}{0ex}}\Rightarrow \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=t/2=20\mathrm{h}/2=10\mathrm{h}\phantom{\rule{0ex}{0ex}}$

100 hours after the beginning,

$\mathrm{Count}\mathrm{rate},A"=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow A"=\frac{4\times {10}^{6}}{{2}^{100/10}}\phantom{\rule{0ex}{0ex}}=0.390625\times {10}^{4}\phantom{\rule{0ex}{0ex}}=3.9\times {10}^{3}\mathrm{disintegrations}/\mathrm{sec}$

#### Page No 443:

#### Question 25:

The half-life of ^{226}Ra is 1602 y. Calculate the activity of 0.1 g of RaCl_{2} in which all the radium is in the form of ^{226}Ra. Taken atomic weight of Ra to be 226 g mol^{−1} and that of Cl to be 35.5 g mol^{−1}.

#### Answer:

Given:

Half-life of radium, *T*_{1}_{/2} = 1602 years

Atomic weight of radium = 226 g/mole

Atomic weight of chlorine = 35.5 g/mole

Now,

1 mole of RaCl_{2} = 226 + 71 = 297 g

297 g = 1 mole of RaCl_{2}

0.1 g = $\frac{1}{297}\times 0.1\mathrm{mole}\mathrm{of}{\mathrm{RaCl}}_{2}$

Total number of atoms in 0.1 g of RaCl_{2}, *N *$=\frac{0.1\times 6.023\times {10}^{23}}{297}=0.02027\times {10}^{22}\phantom{\rule{0ex}{0ex}}$^{}

∴ No of atoms, *N* = 0.02027 $\times $ 10^{22}

$\mathrm{Disintegration}\mathrm{constant},\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\phantom{\rule{0ex}{0ex}}=\frac{0.693}{1602\times 365\times 24\times 3600}\phantom{\rule{0ex}{0ex}}=1.371\times {10}^{-11}$

Activity of radioactive sample, *A* = $\lambda N$

= $1.371\times {10}^{-11}\times 2.027\times {10}^{20}$

= $2.8\times {10}^{9}\mathrm{disintegrations}/\mathrm{second}$

#### Page No 443:

#### Question 26:

The half-life of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.

#### Answer:

Given:

Half-life of radioisotope, ${T}_{1/2}$ = 10 hrs

Initial activity, *A*_{0} = 1 Ci

Disintegration constant, $\lambda =\frac{0.693}{10\times 3600}{\mathrm{s}}^{-1}$

Activity of radioactive sample, $A={A}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}$^{}

Here,* **A*_{0} = Initial activity

$\lambda $ = Disintegration constant

*t* = Time taken

After 9 hours,

Activity, $A={A}_{0}{e}^{-\lambda t}=1\times {e}^{-\frac{0.693}{10\times 3600}\times 9}=0.536\mathrm{Ci}$

∴ Number of atoms left, *N* = $\frac{A}{\lambda}$ = $\frac{0.536\times 10\times 3.7\times {10}^{10}\times 3600}{0.693}=103.023\times {10}^{13}$

^{}

$Activity,A\mathit{"}={A}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}=1\times {e}^{\frac{-0.693}{10}\times 10}=0.5\mathrm{Ci}$

Number of atoms left after the 10^{th} hour $\left(N"\right)$ will be

$A"=\lambda N"\phantom{\rule{0ex}{0ex}}N"=\frac{A"}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{0.5\times 3.7\times {10}^{10}\times 3.600}{0.693/10}\phantom{\rule{0ex}{0ex}}=26.37\times {10}^{10}\times 3600=96.103\times {10}^{13}$

Number of disintegrations = (103.023 − 96.103) × 10^{13}

= 6.92 × 10^{13}

#### Page No 443:

#### Question 27:

The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old ^{32}P(*t*_{1}_{/2} = 14.3 days) source if it was originally purchased for 800 rupees?

#### Answer:

Given:

Half-life of ^{32}P source, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 14.3 days

Time, *t* = 30 days = 1 month

Here, the selling rate of a radioactive isotope is decided by its activity.

∴ Selling rate = Activity of the radioactive isotope after 1 month

Initial activity,* **A*_{0} = 800 disintegration/sec

Disintegration constant $\left(\lambda \right)$ is given by

$\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{14.3}{\mathrm{days}}^{-1}$

Activity $\left(A\right)$ is given by

*A* = *A*_{0}*e*^{−}^{λt}

Here, $\lambda $ = Disintegration constant

$\therefore $ Activity of the radioactive isotope after one month (selling rate of the radioactive isotope) $\left(A\right)$ is given below. * *

$A=800\times {e}^{\frac{-0.693}{14.3}\times 30}\phantom{\rule{0ex}{0ex}}=800\times 0.233669\phantom{\rule{0ex}{0ex}}=186.935=\mathrm{Rs}187$

#### Page No 443:

#### Question 28:

^{57}Co decays to ^{57}Fe by β^{+}- emission. The resulting ^{57}Fe is in its excited state and comes to the ground state by emitting γ-rays. The half-life of β^{+}- decay is 270 days and that of the γ-emissions is 10^{−8} s. A sample of ^{57}Co gives 5.0 × 10^{9} gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 10^{9} per second?

#### Answer:

According to the question, when the β^{+} decays to half of its original amount, the emission rate of γ-rays will drop to half. For this, the sample will take 270 days.

Therefore, the required time is 270 days.

#### Page No 443:

#### Question 29:

Carbon (*Z* = 6) with mass number 11 decays to boron (*Z* = 5). (a) Is it a β^{+}-decay or a β^{−}decay? (b) The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?

#### Answer:

(a) The reaction is given by

${\mathrm{C}}_{6}\to {\mathrm{B}}_{5}+{\mathrm{\beta}}^{+}+\mathrm{v}$

It is a β^{+} decay since atomic number is reduced by 1.

(b) Half-life of the decay scheme, *T*_{1/2 } = 20.3 minutes

Disintegration constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{20.3}{\mathrm{min}}^{-1}$

If* t* is the time taken by the mixture in converting, let the total no. of atoms be 100*N*_{0}_{.}

Carbon | Boron | |

Initial | 90 N_{0} |
10 N_{0} |

Final | 10 N_{0} |
90 N_{0} |

*N*=

*N*

_{0}e

^{−λ}

^{t}

Here,

*N*

_{0}= Initial number of atoms

*N*= Number of atoms left undecayed

10

*N*

_{0}=

_{ }90

*N*

_{0}e

^{−λ}

^{t}( For carbon)

$\Rightarrow \frac{1}{9}={e}^{-\frac{0.693}{20.3}\times t}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{In}\frac{1}{9}=\frac{-0.693}{20.3}t\phantom{\rule{0ex}{0ex}}\Rightarrow t=64.36=64\mathrm{min}$

#### Page No 443:

#### Question 30:

4 × 10^{23} tritium atoms are contained in a vessel. The half-life of decay tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decay in the next 10 hours (c) the number of decays in the next 6.15 y.

#### Answer:

Given:

Number of tritium atoms, *N _{0}* = 4 × 10

^{23}

Half-life of tritium nuclei, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 12.3 years

Disintegration constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{0.693}{12.3}$ years

^{$-1$}

Activity of the sample $\left(A\right)$ is given by

*A*

_{0}= $\frac{dN}{dt}$ = $\lambda {N}_{0}$

$\Rightarrow {A}_{0}=\frac{0.693}{{t}_{1/2}}{N}_{0}\phantom{\rule{0ex}{0ex}}=\frac{0.693}{12.3}\times 4\times {10}^{23}\mathrm{disintegration}/\mathrm{year}\phantom{\rule{0ex}{0ex}}=\frac{0.693\times 4\times {10}^{23}}{12.3\times 3600\times 24\times 365}\mathrm{disintegration}/\mathrm{sec}\phantom{\rule{0ex}{0ex}}=7.146\times {10}^{14}\mathrm{disintegration}/\mathrm{sec}\phantom{\rule{0ex}{0ex}}$

(b) Activity of the sample,

*A*= 7.146$\times $10

^{14}disintegration/sec

$\mathrm{Number}\mathrm{of}\mathrm{decays}\mathrm{in}\mathrm{the}\mathrm{next}10\mathrm{h}\mathrm{ours}=7.146\times {10}^{14}\times 10\times 3600\phantom{\rule{0ex}{0ex}}=257.256\times {10}^{17}\phantom{\rule{0ex}{0ex}}=2.57\times {10}^{19}\phantom{\rule{0ex}{0ex}}$

(c) Number of atoms left undecayed, $N={N}_{0}{e}^{-\lambda t}$

Here,

*N*

_{0}= Initial number of atoms

$\therefore N=4\times {10}^{23}\times {e}^{\frac{-0.693}{12.3}\times 6.15}=2.83\times {10}^{23}$

Number of atoms disintegrated = (

*N*

_{0}

_{ $-$ }

*N*) = (4$-$2.83) × 10

^{23}= 1.17 × 10

^{23}

#### Page No 443:

#### Question 31:

A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm^{2} window. If the source contains 6.0 × 10^{16} active nuclei and the counter records a rate of 50000 counts/second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.

#### Answer:

Given:

Counts received per second = 50000 Counts/second

Number of active nuclei, *N* = 6 × 10^{16}

Total counts radiated from the source, $\frac{dN}{\mathrm{dt}}$ = Total surface area × 50000 counts/cm^{2}

= 4 × 3.14 × 1 × 10^{4} × 5 × 10^{4}

= 6.28 × 10^{9} Counts

We know

$\frac{dN}{dt}=\lambda N$

Here, *λ =* Disintegration constant

$\therefore \lambda =\frac{6.28\times {10}^{9}}{6\times {10}^{16}}\phantom{\rule{0ex}{0ex}}=1.0467\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=1.05\times {10}^{-7}{\mathrm{s}}^{-1}$

#### Page No 443:

#### Question 32:

^{238}U decays to ^{206}Pb with a half-life of 4.47 × 10^{9} y. This happens in a number of steps. Can you justify a single half for this chain of processes? A sample of rock is found to contain 2.00 mg of ^{238}U and 0.600 mg of ^{206}Pb. Assuming that all the lead has come from uranium, find the life of the rock.

#### Answer:

Given:

Half-life of ^{238}U, *t*_{1}_{/2} = 4.47 × 10^{9} years^{}

$\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{atoms}\mathrm{present}\mathrm{in}\mathrm{the}\mathrm{rock}initially,{N}_{0}=\left(\frac{6.023\times {10}^{23}\times 2}{238}+\frac{6.023\times {10}^{23}\times 0.6}{206}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{12.046}{238}+\frac{3.62}{206}\right)\times {10}^{20}\phantom{\rule{0ex}{0ex}}=\left(0.0505+0.0175\right)\times {10}^{20}\phantom{\rule{0ex}{0ex}}=0.0680\times {10}^{20}\phantom{\rule{0ex}{0ex}}$

Now, *N* = *N*_{0}e^{$-\lambda t$}

Here, $\lambda $ = Disintegration constant

*t* = Life of the rock

$\Rightarrow N={N}_{0}{e}^{\frac{-0.693}{{t}_{1/2}}\times t}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.0505=0.0680{e}^{\frac{-0.693}{4.47\times {10}^{9}}\times t}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ln}\left(\frac{0.0505}{0.0680}\right)=\frac{-0.6931}{4.47\times {10}^{9}}\times t\phantom{\rule{0ex}{0ex}}\Rightarrow t=1.92\times {10}^{9}\mathrm{years}$

#### Page No 443:

#### Question 33:

When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ^{14}C per gram per minute. A sample from an ancient piece of charcoal shows ^{14}C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of ^{14}C is 5730 y.

#### Answer:

Given:

Initial activity of charcoal, *A*_{0} = 15.3 disintegrations per gram per minute

Half-life of charcoal, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 5730 years

Final activity of charcoal after a few years,* A = *12.3 disintegrations per gram per minute

Disintegration constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{5370}{\mathrm{y}}^{-1}$

Let the sample take a time of* t* years for the activity to reach 12.3 disintegrations per gram per minute.

Activity of the sample, $A={A}_{0}{e}^{-\lambda t}$

$A={A}_{0}{e}^{-\frac{0.693}{5730}\times t}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{In}\frac{12.3}{15.3}=\frac{-0.693}{5730}t\phantom{\rule{0ex}{0ex}}\Rightarrow 0.218253=\frac{0.693}{5730}\times t\phantom{\rule{0ex}{0ex}}\Rightarrow t=1804.3\mathrm{years}$

#### Page No 443:

#### Question 34:

Natural water contains a small amount of tritium (${}_{1}{}^{3}\mathrm{H}$). This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottled of whisky. On returning, he analyses the whisky and finds that it contains only 1.5 per cent of the ${}_{1}{}^{3}\mathrm{H}$ radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt.

#### Answer:

Given:

Half-life time of tritium, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 12.5 years

Disintegration constant, $\lambda =\frac{0.693}{12.5}\mathrm{per}\mathrm{year}$

Let* **A*_{0} be the activity, when the bottle was manufactured.

Activity after 8 years $\left(A\right)$ is given by

$A={A}_{0}{e}^{\frac{-0.693}{12.5}\times 8}$ ...(1)

Let us consider that the mountaineering had taken place *t *years ago.

Then, activity of the bottle $\left(A\text{'}\right)$ on the mountain is given by

$A\text{'}={A}_{0}{e}^{-\lambda t}$

Here, *A'* = (Activity of the bottle manufactured 8 years ago) × 1.5%

$A\text{'}={A}_{0}{e}^{\frac{-0.693}{12.5}\times 8}\times 0.015...\left(2\right)\phantom{\rule{0ex}{0ex}}$

Comparing (1) and (2)

$\frac{0.693}{12.5}t=\frac{-0.6931\times 8}{12.5}+\mathrm{In}[0.015]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-0.693}{12.5}t=\frac{-0.693}{12.5}\times 8-4.1997\phantom{\rule{0ex}{0ex}}\Rightarrow 0.693t=58.040\phantom{\rule{0ex}{0ex}}\Rightarrow t=83.75\mathrm{years}$

#### Page No 443:

#### Question 35:

The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing ^{128}I varies with time as follows.

Time t (minute): |
0 | 25 | 50 | 75 | 100 |

Ctount rate R (10^{9} s^{−1}): |
30 | 16 | 8.0 | 3.8 | 2.0 |

*R*

_{0}/

*R*) against

*t*. (b) From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the half-life

*t*

_{1}

_{/2}.

#### Answer:

(a) For *t* = 0,

$\mathrm{ln}\left(\frac{{R}_{\mathit{0}}}{R}\right)=\mathrm{In}\left(\frac{30\times {10}^{9}}{30\times {10}^{9}}\right)=0\phantom{\rule{0ex}{0ex}}$

For *t* = 25 s,

$\mathrm{ln}\left(\frac{{R}_{\mathit{0}}}{{R}_{\mathit{2}}}\right)=\mathrm{In}\left(\frac{30\times {10}^{9}}{16\times {10}^{9}}\right)=0.63$

For *t *= 50 s,

$\mathrm{In}\left(\frac{{\mathit{R}}_{\mathit{0}}}{{\mathit{R}}_{\mathit{3}}}\right)=\mathrm{In}\left(\frac{30\times {10}^{9}}{8\times {10}^{9}}\right)=1.35$

For *t* = 75 s,

$\mathrm{ln}\left(\frac{{R}_{0}}{{R}_{4}}\right)=\mathrm{In}\left(\frac{30\times {10}^{9}}{3.8\times {10}^{9}}\right)=2.06$

For *t* = 100 s,

$\mathrm{In}\left(\frac{{R}_{0}}{{R}_{5}}\right)=\mathrm{In}\left(\frac{30\times {10}^{9}}{2\times {10}^{9}}\right)=2.7$

The required graph is shown below.

(b) Slope of the graph = 0.028

∴ Decay constant, $\lambda $ = 0.028 min^{$-1$}

The half-life period$\left({T}_{\frac{1}{2}}\right)$ is given by

${T}_{\frac{1}{2}}=\frac{0.693}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{0.693}{0.028}=25\mathrm{min}$

#### Page No 443:

#### Question 36:

The half-life of ^{40}K is 1.30 × 10^{9} y. A sample of 1.00 g of pure KCI gives 160 counts s^{−1}. Calculate the relative abundance of ^{40}K (fraction of ^{40}K present) in natural potassium.

#### Answer:

Given:

Half-life period of ^{40}K, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 1.30 × 10^{9} years

Count given by 1 g of pure KCI, *A *= 160 counts/s

Disintegration constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Now, activity, *A* =* **λN*

$\Rightarrow 160=\frac{0.693}{{t}_{1/2}}\times N\phantom{\rule{0ex}{0ex}}\Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^{9}\times 365\times 86400}\right)\times N\phantom{\rule{0ex}{0ex}}\Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^{9}}{0.693}\phantom{\rule{0ex}{0ex}}\Rightarrow N=9.5\times {10}^{18}$^{}

6.023 × 10^{23} atoms are present in 40 gm.

$\mathrm{Thus,}\mathrm{9.5\; \times \; 1018}\mathrm{atoms}\mathrm{will}\mathrm{be}\mathrm{present}in\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}}\mathrm{gm}.\phantom{\rule{0ex}{0ex}}=\frac{4\times 9.5\times {10}^{-4}}{6.023}\mathrm{gm}\phantom{\rule{0ex}{0ex}}=6.309\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=0.00063\mathrm{gm}$

Relative abundance of ^{40}K in natural potassium = (2 × 0.00063 × 100)% = 0.12%

#### Page No 443:

#### Question 37:

${}_{80}{}^{197}\mathrm{Hg}$ decay to ${}_{79}{}^{197}\mathrm{Au}$ through electron capture with a decay constant of 0.257 per day. (a) What other particle or particles are emitted in the decay? (b) Assume that the electron is captured from the K shell. Use Moseley's law √*v* = *a*(*Z* − *b*) with *a* = 4.95 × 10^{7} s^{−1}^{/2} and *b* = 1 to find the wavelength of the K_{α} X-ray emitted following the electron capture.

#### Answer:

Given:

Decay constant of electron capture = 0.257 per day

(a) The reaction is given as

${}_{80}{}^{197}\mathrm{Hg}+\mathrm{e}\to {}_{79}{}^{197}\mathrm{Au}+\mathrm{v}$

The other particle emitted in this reaction is neutrino v*.*

(b) Moseley's law is given by

$\sqrt{v}$ = *a*(*Z* − *b*)

We know

$v=\frac{c}{\lambda}$

Here, *c* = Speed of light

$\lambda $ = Wavelength of the K_{α} X-ray

$\sqrt{\frac{c}{\lambda}}=4.95\times {10}^{7}(79-1)\phantom{\rule{0ex}{0ex}}=4.95\times {10}^{7}\times 78\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{c}{\lambda}=(4.95\times 78{)}^{2}\times {10}^{14}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{3\times {10}^{8}}{149073.21\times {10}^{14}}\phantom{\rule{0ex}{0ex}}=20\mathrm{pm}$

#### Page No 443:

#### Question 38:

A radioactive isotope is being produced at a constant rate *dN/dt* = *R* in an experiment. The isotope has a half-life *t*_{1}_{/2}. Show that after a time *t *>> *t*_{1}_{/2} the number of active nuclei will become constant. Find the value of this constant.

#### Answer:

Given:

Half life period of isotope = *t*_{1}_{/2}

Disintegration constant, $\lambda =\frac{0.693}{{t}_{{\displaystyle 1/2}}}$

Rate of Radio active decay $\left(R\right)$ is given by,

$R=\frac{d\mathrm{N}}{dt}$

We are to show that after time *t* >> *t*_{1}_{/2} the number of active nuclei is constant.

${\left(\frac{dN}{dt}\right)}_{Present}=R={\left(\frac{dN}{dt}\right)}_{decay}\phantom{\rule{0ex}{0ex}}\therefore R={\left(\frac{dN}{dt}\right)}_{decay}\phantom{\rule{0ex}{0ex}}$

Rate of radioactive decay, $R=\lambda N$

Here, *λ *= Radioactive decay constant

*N* = Constant number

$R=\frac{0.693}{{t}_{1/2}}\times N\phantom{\rule{0ex}{0ex}}\Rightarrow R{t}_{1/2}=0.693N\phantom{\rule{0ex}{0ex}}\Rightarrow N=\frac{R{t}_{1/2}}{0.693}$

This value of *N* should be constant.

#### Page No 443:

#### Question 39:

Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at *t* = 0. Find the number of active nuclei at time *t*.

#### Answer:

Let the number of atoms present at *t* = 0 be *N*_{0}.

Let *N* be the number of radio-active isotopes present at time *t.*

Then,

*N *=* **N*_{0}*e*^{−λt}

Here, $\lambda $ = Disintegration constant

∴ Number of radioactive isotopes decayed = *N*_{0} − *N* = *N*_{0} − *N*_{0}*e*^{−}^{λt}

=* **N*_{0} (1−*e*^{−}^{λt}) ...(1)

Rate of decay $\left(R\right)$ is given by

* R* = *λN _{0}* ...(2)

Substituting the value of

*N*

_{0}from equation (2) to equation (1), we get

$N={N}_{0}(1-{e}^{-\mathrm{\lambda}t})\phantom{\rule{0ex}{0ex}}=\frac{R}{\lambda}(1-{e}^{-\mathrm{\lambda}t})$

#### Page No 443:

#### Question 40:

In an agricultural experiment, a solution containing 1 mole of a radioactive material (*t*_{1}_{/2} = 14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruit. If the activity measured was 1 µCi, what per cent of activity is transmitted from the root to the fruit in steady state?

#### Answer:

Given:

Initial no of atoms, *N*_{0} = 1 mole = 6 × 10^{23} atoms

Half-life of the radioactive material, *T*_{1}_{/2} = 14.3 days

Time taken by the plant to settle down,* t* = 70 h

Disintegration constant, $\lambda =\frac{0.693}{{t}_{1/2}}$ = $\frac{0.693}{14.3\times 24}$ h^{$-$1}

*N* = *N*_{0}*e*^{−λ}^{t}

$=6\times {10}^{23}\times {e}^{\frac{-0.693\times 70}{14.3\times 24}}\phantom{\rule{0ex}{0ex}}=6\times {10}^{23}\times 0.868\phantom{\rule{0ex}{0ex}}=5.209\times {10}^{23}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Activity,}R=\frac{dN}{dt}=5.209\times {10}^{23}\times \frac{0.693}{14.3\times 24}\phantom{\rule{0ex}{0ex}}=\frac{0.0105\times {10}^{23}}{3600}\mathrm{dis}/\mathrm{hr}\phantom{\rule{0ex}{0ex}}=2.9\times {10}^{-6}\times {10}^{23}\mathrm{dis}/\mathrm{sec}\phantom{\rule{0ex}{0ex}}=2.9\times {10}^{17}\mathrm{dis}/\mathrm{sec}$

$\mathrm{Fraction}\mathrm{of}\mathrm{activity}\mathrm{transmitted}=\left(\frac{1\mathrm{\mu Ci}}{2.9\times {10}^{17}}\right)\times 100\%\phantom{\rule{0ex}{0ex}}=\left[\left(\frac{1\times 3.7\times {10}^{4}}{2.9\times {10}^{17}}\right)\times 100\right]\%\phantom{\rule{0ex}{0ex}}=1.275\times {10}^{-11}\%$

#### Page No 443:

#### Question 41:

A vessel of volume 125 cm^{3} contains tritium (^{3}H, *t*_{1}_{/2} = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.

#### Answer:

Given:

Volume of the vessel, *V *= 125 cm^{3} = 0.125 L

Half-life time of tritium, *t*_{1}_{/2} = 12.3 y = 3.82 × 10^{8} s

Pressure, *P *= 500 kpa = 5 atm

Temperature,* T* = 300 K,

Disintegration constant, $\lambda =\frac{0.693}{{t}_{1/2}}$

=$\phantom{\rule{0ex}{0ex}}\frac{0.693}{3.82\times {10}^{8}}=0.1814\times {10}^{-8}\phantom{\rule{0ex}{0ex}}=1.81\times {10}^{-9}{\mathrm{s}}^{-1}$

No. of atoms left undecayed, *N* = *n* × 6.023 × 10^{23}

$=\frac{5\times 0.125}{8.2\times {10}^{-2}\times 3\times {10}^{2}}\times 6.023\times {10}^{23}\left(\because n=\frac{PV}{RT}\right)\phantom{\rule{0ex}{0ex}}=1.5\times {10}^{22}\mathrm{atoms}\phantom{\rule{0ex}{0ex}}$

Activity, *A* = *λN*

= 1.81 × 10^{−9} × 1.5 × 10^{22} = 2.7 × 10^{13} disintegration/sec

$\therefore A=\frac{2.7\times {10}^{13}}{3.7\times {10}^{10}}\mathrm{Ci}=729.72\mathrm{Ci}$

#### Page No 444:

#### Question 42:

${}_{83}{}^{212}\mathrm{Bi}$ can disintegrate either by emitting an α-particle of by emitting a β^{−}-particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration α-and β-decays are in the ratio 7/13. The overall half-life of ^{212}Bi is one hour. If 1 g of pure ^{212}Bi is taken at 12.00 noon, what will be the composition of this sample at 1 P.m. the same day?

#### Answer:

Given:

Half-life of ^{212}Bi, *T*_{1}_{/2} = 1 h^{$-1$}

When ${}_{83}{}^{212}\mathrm{Bi}$ disintegrates by emitting an α-particle

${}_{83}{}^{212}\mathrm{Bi}\to {}_{81}{}^{208}\mathrm{T}1+{}_{2}{}^{4}\mathrm{He}\left(\mathrm{\alpha}\right)\phantom{\rule{0ex}{0ex}}$

When ${}_{83}{}^{212}\mathrm{Bi}$ disintegrates by emitting a β^{−}particle

$\phantom{\rule{0ex}{0ex}}{}_{83}{}^{212}\mathrm{Bi}\to {}_{84}{}^{212}\mathrm{P}_{0}+{\mathrm{\beta}}^{-}+\overline{\mathrm{v}}\phantom{\rule{0ex}{0ex}}$

Half-life period of ^{212}Bi, ${T}_{\frac{1}{2}}$= 1 h^{$-1$}

At *t* = 0, the amount of ^{212}Bi present = 1 g

At *t* = 1 = One half-life,

Amount of ^{212}Bi present = 0.5 g

Probability of disintegration of α-decay and β-decay are in the ratio $\frac{7}{13}$.

In 20 g of ^{212}Bi, the amount of ^{208}Ti formed = 7 g

In 1 g of ^{212}Bi, the amount of ^{208}Ti formed = 7/20 g

∴ Amount of ^{208}Ti present in 0.5 g = $\frac{7}{20}\times 0.5=0.175\mathrm{g}$

In 20 g of ^{212}Bi, the amount of ^{212}Po formed = 13 g

In 1 g of ^{212}Bi, the amount of ^{212}Po formed = 13/20 g

∴ Amount of ^{212}Po present in 0.5 g = $\frac{13}{20}\times 0.5=0.325\mathrm{g}$

#### Page No 444:

#### Question 43:

A sample contains a mixture of ^{108}Ag and ^{110}Ag isotopes each having an activity of 8.0 × 10^{8} disintegration per second. ^{110}Ag is known to have larger half-life than ^{108}Ag. The activity *A* is measured as a function of time and the following data are obtained.

Time (s) |
Activity (A) (10 ^{8} disinte-grations s ^{−1}) |
Time (s) |
Activity (A 10 ^{8} disinte-grations s^{−1}) |

20 40 60 80 100 |
11.799 9.1680 7.4492 6.2684 5.4115 |
200 300 400 500 |
3.0828 1.8899 1.1671 0.7212 |

(a) Plot ln (

*A/*

*A*

_{0}) versus time. (b) See that for large values of time, the plot is nearly linear. Deduce the half-life of

^{110}Ag from this portion of the plot. (c) Use the half-life of

^{110}Ag to calculate the activity corresponding to

^{108}Ag in the first 50 s. (d) Plot In (A/A

_{0}) versus time for

^{108}Ag for the first 50 s. (e) Find the half-life of

^{108}Ag.

#### Answer:

(a) Activity, *A*_{0} = 8 × 10^{8} dis/sec

(i) $\mathrm{In}\left(\frac{{A}_{1}}{{A}_{0}}\right)=\mathrm{In}\left(\frac{11.794}{8}\right)=0.389\phantom{\rule{0ex}{0ex}}$

(ii) $\mathrm{In}\left(\frac{{A}_{2}}{{A}_{0}}\right)=\mathrm{In}\left(\frac{9.1680}{8}\right)=0.12362$

(iii) $\mathrm{In}\left(\frac{{A}_{3}}{{A}_{0}}\right)=\mathrm{In}\left(\frac{7.4492}{8}\right)=-0.072$

(iv) $\mathrm{In}\left(\frac{{A}_{4}}{{A}_{0}}\right)=\mathrm{In}\left(\frac{6.2684}{6}\right)=-0.244\phantom{\rule{0ex}{0ex}}$

(v) $\mathrm{In}\left(\frac{{A}_{5}}{A}\right)$ = $\mathrm{In}\left(\frac{5.4115}{8}\right)=-0.391\phantom{\rule{0ex}{0ex}}$

(vi) $\mathrm{In}\left(\frac{{\mathrm{A}}_{6}}{{\mathrm{A}}_{0}}\right)$ = $\mathrm{In}\left(\frac{3.0828}{8}\right)=-0.954$

(vii)$\mathrm{In}\left(\frac{{\mathrm{A}}_{7}}{{\mathrm{A}}_{0}}\right)$= $\mathrm{In}\left(\frac{91.8899}{8}\right)=-1.443\phantom{\rule{0ex}{0ex}}$

(viii) $\mathrm{In}\left(\frac{1.1671}{8}\right)=\mathrm{In}\left(\frac{90.7212}{8}\right)=-1.93\phantom{\rule{0ex}{0ex}}$

(ix) $\mathrm{In}\left(\frac{0.7212}{8}\right)=\mathrm{In}\left(\frac{90.7212}{8}\right)=-2.406$

The required graph is given below.

(b) Half-life of ^{110}Ag = 24.4 s

(c) Half-life of ^{110}Ag, ${T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 24.4 s

Decay constant, $\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$\Rightarrow $ $\lambda =\frac{0.693}{24.4}=0.0284$

∴ *t* = 50 sec

$\mathrm{Activity},\mathrm{A}={\mathrm{A}}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}=8\times {10}^{8}\times {e}^{-0.0284\times 50}\phantom{\rule{0ex}{0ex}}=1.93\times {10}^{8}$

(d)

(e) The half-life period of ^{108}Ag that you can easily watch in your graph is 144 s.

#### Page No 444:

#### Question 44:

A human body excretes (removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing ^{99}Tc. This isotope is radioactive with a half-life of 6 hours. The activity from the body just after the injection is 6 μCi. How much time will elapse before the activity falls to 3 μCi?

#### Answer:

Given:

Time taken by the body to excrete half the amount, *t*_{1} = 24 hours

Half-life of radioactive isotope, *t*_{2} = 6 hours

Initial activity, *A*_{0} = 6 μCi

Let after time *t*, activity of the sample be *A*.

Half-life period $\left({T}_{1/2}\right)$ is given by

${T}_{1/2}=\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}=\frac{24\times 6}{24+6}\phantom{\rule{0ex}{0ex}}=\frac{24\times 6}{30}=4.8\mathrm{h}$

Activity $\left(A\right)$ at time *t* is given by

$\therefore \mathrm{A}=\frac{{\mathrm{A}}_{0}}{{2}^{t/{T}_{1/2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 3\mathrm{\mu C}i=\frac{6\mathrm{\mu Ci}}{{2}^{t/4.8}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6\mathrm{\mu Ci}}{{2}^{t/4.8}}=3\phantom{\rule{0ex}{0ex}}\Rightarrow t=4.8\mathrm{h}$

#### Page No 444:

#### Question 45:

A charged capacitor of capacitance *C* is discharged through a resistance *R*. A radioactive sample decays with an average-life *τ*. Find the value of *R* for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

#### Answer:

Discharging of a capacitor through a resistance *R* is given by

$Q=q{e}^{-t/CR}$

Here, *Q *= Charge left

*q* = Initial charge

*C* = Capacitance

*R *= Resistance

Energy, *E* = $\frac{1}{2}\frac{{Q}^{2}}{C}$ = $\frac{{q}^{2}{e}^{{\displaystyle -2t/CR}}}{2C}$

Activity, *A* =* **A*_{0}e^{$-\lambda t$}

Here, A_{0} = Initial activity

$\lambda $ = Disintegration constant

∴ Ratio of the energy to the activity = $\frac{E}{A}=\frac{{q}^{2}\times {e}^{-2t/CR}}{2C{A}_{0}{e}^{-\lambda t}}$

Since the terms are independent of time, their coefficients can be equated.

$\frac{2t}{CR}=\lambda t$

$\Rightarrow \lambda =\frac{2}{CR}$

$\Rightarrow \frac{1}{\tau}=\frac{2}{CR}$

$\Rightarrow R=2\frac{\tau}{C}$

#### Page No 444:

#### Question 46:

Radioactive isotopes are produced in a nuclear physics experiment at a constant rate *dN/dt* = *R*. An inductor of inductance 100 mH, a resistor of resistance 100 Ω and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that *i/N* remains constant in time where *i* is the current in the circuit at time *t* and *N* is the number of active nuclei at time *t*. Find the half-life of the isotope.

#### Answer:

Given:

Resistance of resistor, *R = *100 Ω

Inductance of an inductor, *L* = 100 mH

Current $\left(i\right)$ at any time $\left(t\right)$ is given by

$i={i}_{0}\left(1-{e}^{\frac{-Rt}{L}}\right)$

Number of active nuclei $\left(N\right)$ at any time $\left(t\right)$ is given by

$N={N}_{0}{e}^{-\lambda t}$

Where *N*_{0} = Total number of nuclei

$\lambda $ = Disintegration constant

Now,

$\frac{i}{N}=\frac{{i}_{0}\left(1-{e}^{-tR/L}\right)}{{N}_{0}{e}^{-\lambda t}}$

As $\frac{i}{N}$ is independent of time, coefficients of *t* are equal.

Let ${t}_{\frac{1}{2}}$ be the half-life of the isotope.

$\Rightarrow \frac{-R}{L}=-\lambda \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{R}{L}=\frac{0.693}{{t}_{{\displaystyle \frac{1}{2}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{\frac{1}{2}}=0.693\times {10}^{-3}=6.93\times {10}^{-4}\mathrm{s}$

#### Page No 444:

#### Question 47:

Calculate the energy released by 1g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope ^{235}U has an abundance of 0.7% by weight in natural uranium.

#### Answer:

235 g of uranium contains 6.02 × 10^{23} atoms.

1 g of uranium = $\frac{1}{235}\times 6.023\times {10}^{23}$ atoms

∴ 0.7 g of uranium = $\frac{1}{235}\times 6.023\times {10}^{23}\times 0.007$ atoms

1 atom gives 200 MeV.

∴ Total energy released = $\frac{6.023\times {10}^{23}\times 0.007\times 200\times {10}^{6}\times 1.6\times {10}^{-19}}{235}\mathrm{J}$ = 5.74$\times $10^{8 }J

#### Page No 444:

#### Question 48:

A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of ^{235}U being consumed every second. Average released per fission is 200 MeV.

#### Answer:

Given:

Rate of development of thermal energy = 300 MW

Average energy released per fission = 200 MeV

Let *N* be the number of atoms disintegrating per second.

Then, the total energy emitted per second will be

*N$\times $*200$\times $10^{6}$\times $1.6$\times $10^{$-19$} = Power

*N$\times $*200$\times $10^{6}$\times $1.6$\times $10^{$-19$} = 300$\times $10^{6}

$\Rightarrow N=\frac{3}{2\times 1.6}\times {10}^{19}=\frac{3}{3.2}\times {10}^{19}$ atoms

6.023 × 10^{23} atoms = 238 gm of U^{235}

$\frac{3}{3.2}\times {10}^{19}\mathrm{atoms}\mathrm{will}\mathrm{present}\mathrm{in}\frac{238\times 3\times {10}^{19}}{6\times {10}^{23}\times 3.2}=3.7\mathrm{mg}$

#### Page No 444:

#### Question 49:

A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through ^{235}U, at what rate will the amount of ^{235}U decrease? Express your answer in kg per day. (c) Assuming that uranium enriched to 3% in ^{235}U will be used, how much uranium is needed per month (30 days)?

#### Answer:

(a) Total population of the town = 1 million = 10^{6 }

Average electric power needed per person = 300 W

Total power used by the town in one day = 300 × 10^{6} × 60 × 60 × 24 J = 300 × 86400 ×10^{6} J

Energy generated in one fission = 200 × 10^{6} × 1.6 × 10^{−19} J =3.2 × 10^{−11} J

The efficiency with which thermal power is converted into electric power is 25%.

Therefore, Electrical energy is given by

$\therefore \mathrm{Electrical}\mathrm{energy},\mathrm{E}=3.2\times {10}^{-11}\times \frac{25}{100}\phantom{\rule{0ex}{0ex}}\mathrm{E}=8\times {10}^{-12}\mathrm{J}$

Let the number of fission be *N*.

So, total energy of *N* fissions = *N* × 8 × 10^{−12}

As per the question,

*N* × 8 × 10^{−12} = 300 × 86400 × 10^{6} J

*N* = 3.24 × 10^{24}

(b) Number of moles required per day *n* = $\frac{N}{6.023\times {10}^{23}}$

$\Rightarrow n=\frac{3.24\times {10}^{24}}{6.023\times {10}^{23}}=5.38\mathrm{mol}$

So, the amount of uranium required per day = 5.38 × 235

= 1264.3 gm = 1.2643 kg

(c) Total uranium needed per month = 1.264 × 30 kg

Let *x* kg of uranium enriched to 3% in ^{235}U be used.

$\Rightarrow x\times \frac{3}{100}=1.264\times 30\phantom{\rule{0ex}{0ex}}\Rightarrow x=1264\mathrm{kg}$

#### Page No 444:

#### Question 50:

Calculate the *Q*-values of the following fusion reactions:

(a) ${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{2}\mathrm{H}\to {}_{1}{}^{3}\mathrm{H}+{}_{1}{}^{1}\mathrm{H}$

(b) ${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{2}\mathrm{H}\to {}_{2}{}^{3}\mathrm{He}+n$

(c) ${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{3}\mathrm{H}\to {}_{2}{}^{4}\mathrm{He}+n$.

Atomic masses are

$m\left({}_{1}{}^{2}H\right)=2.014102\mathrm{u},\phantom{\rule{0ex}{0ex}}m\left({}_{1}{}^{3}H\right)=3.016049\mathrm{u},\phantom{\rule{0ex}{0ex}}m\left({}_{2}{}^{3}He\right)=3.016029\mathrm{u},\phantom{\rule{0ex}{0ex}}m\left({}_{2}{}^{4}He\right)=4.002603\mathrm{u}.$

#### Answer:

$\left(\mathrm{a}\right)Q=\left[2\times m\mathit{}{}_{1}{}^{2}\mathrm{H}-\left(m\left({}_{3}{}^{3}H\right)+m\left({}_{1}{}^{3}H\right)\right)\right]{\mathrm{c}}^{2}\phantom{\rule{0ex}{0ex}}=(4.028204-4.023874)\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=4.05\mathrm{MeV}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)Q=\left[2\times m{}_{1}{}^{2}\mathrm{H}-(m{}_{2}{}^{3}H+{m}_{n})\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=[4.028204-4.024694)\times 931\phantom{\rule{0ex}{0ex}}=0.00351\times 931\phantom{\rule{0ex}{0ex}}=3.25\mathrm{MeV}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)Q=\left[m{}_{1}{}^{2}\mathrm{H}+m{}_{1}{}^{3}\mathrm{H}-m{}_{1}{}^{4}\mathrm{He}-{m}_{n}\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=(2.014102+3.016049-4.002603-1.008665)\times 931\phantom{\rule{0ex}{0ex}}=17.57\mathrm{MeV}$

#### Page No 444:

#### Question 51:

Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 *kT* equals the Coulomb potential energy at 2 fm.

#### Answer:

Given:

Average thermal energy, *E* = 1.5 *kT*

Point of coulomb potential energy = 2 fm

Potential energy is given by

*U* = $\frac{K{q}_{1}{q}_{2}}{r}$ .....(1)

Here,* K* = $\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}$ = 9$\times $10^{9}

Charge, ${q}_{1}={q}_{2}=2\times 1.6\times {10}^{-19}\mathrm{C}$

Average kinetic energy $\left(E\right)$ is given by

$E=\frac{3}{2}kT$ .....(2)

Here, *k *= Boltzman constant

*T* = Temperature

Equating equation (1) and (2), we get

$\frac{k{q}_{1}{q}_{2}}{r}=\frac{3}{2}kT$

$\Rightarrow T=\frac{2K{q}_{1}{q}_{2}}{3kr}$

= $\frac{2\times 9\times {10}^{9}\times 4\times {\left(1.6\times {10}^{-19}\right)}^{2}}{3\times 1.38\times {10}^{-23}\times 2\times {10}^{-15}}$

= 2.23$\times $10^{10} K

#### Page No 444:

#### Question 52:

Calculate the Q-value of the fusion reaction

^{4}He + ^{4}He = ^{8}Be.

Is such a fusion energetically favourable? Atomic mass of ^{8}Be is 8.0053 u and that of ^{4}He is 4.0026 u.

#### Answer:

Given:

Atomic mass of ^{8}Be = 8.0053 u

Atomic mass of ^{4}He = 4.0026 u

$\mathrm{Required}Q-value=(2\times 4.0026-8.0053){c}^{2}\phantom{\rule{0ex}{0ex}}=0.0001\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=0.0001\times 931\times {10}^{6}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=93.1\mathrm{KeV}$

No, such reaction is not favourable.

#### Page No 444:

#### Question 53:

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction

^{2}H + ^{2}H → ^{3}H + *p*.

Assume that 1.5 × 10^{−2}% of natural water is heavy water D_{2}O (by number of molecules) and all the deuterium is used for fusion.

#### Answer:

^{23 }molecules.

∴ 1000 g of water =$\frac{6.023\times {10}^{23}\times 1000}{18}=3.346\times {10}^{25}$ molecules

% of deuterium = $3.346\times {10}^{25}$ $\times $ $\frac{0.015}{100}$ = 0.05019 $\times $ 10

^{23}

Energy of deuterium = $30.4486\times {10}^{25}$

$=\left[2\times m\left({}^{2}H\right)-m\left({}^{3}H\right)-{m}_{p}\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left(2\times 2.014102\mathrm{u}-3.016049\mathrm{u}-1.007276u\right){c}^{2}\phantom{\rule{0ex}{0ex}}=0.004879\times 931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=4.542349\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=7.262\times {10}^{-13}\mathrm{J}$

Total energy = 0.05019 $\times $ 10

^{23}$\times $ 7.262 $\times $ 10

^{$-$13}J

= 3644 MJ

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