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Question 1:

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.

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Question 2:

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.

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.

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Question 3:

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.

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.

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Question 4:

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.

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.

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Question 5:

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.

Answer:

If we assemble 6 protons and 6 neutrons to form 12C 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 24Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of â€‹24Mg nucleus, more energy is liberated.

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Question 6:

If we assemble 6 protons and 6 neutrons to form 12C 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 24Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of â€‹24Mg nucleus, more energy is liberated.

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. 

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Question 7:

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. 

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) × (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 = Δmc2

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Question 8:

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) × (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 = Δmc2

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.

i.e.    β- decay: np+e+ν¯β+ decay: pn+e++ν

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.

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Question 9:

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.

i.e.    β- decay: np+e+ν¯β+ decay: pn+e++ν

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.

Answer:

It is given that when a boron nucleus (B510) 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,

                 B510+n01X+H24eCharge:     5      0       3       2Mass:        10    1        7      4

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

X=L37i

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Question 10:

It is given that when a boron nucleus (B510) 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,

                 B510+n01X+H24eCharge:     5      0       3       2Mass:        10    1        7      4

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

X=L37i

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.

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Question 11:

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.

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.

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Question 12:

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.

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.

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Question 1:

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.

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 = 112× (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.

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Question 2:

(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 = 112× (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.

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.

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Question 3:

(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.

Answer:


(c) two extra neutrons and no extra electron

12C and 14C 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, â€‹12​C has 6 protons, 6 electrons and 6 neutrons, whereas â€‹14C has 6 electrons, 6 protons and 8 neutrons. 

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Question 4:


(c) two extra neutrons and no extra electron

12C and 14C 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, â€‹12​C has 6 protons, 6 electrons and 6 neutrons, whereas â€‹14C has 6 electrons, 6 protons and 8 neutrons. 

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 1H​1), only protons are present. Due to this, the mass number is equal to the atomic number.

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Question 5:

(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 1H​1), only protons are present. Due to this, the mass number is equal to the atomic number.

Answer:

(a) a straight line
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
R=RoA13RRo=A13lnRRo=13ln A
Therefore, the graph of ln(R/R0) versus ln A is a straight line passing through the origin with slope 1/3.

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Question 6:

(a) a straight line
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
R=RoA13RRo=A13lnRRo=13ln A
Therefore, the graph of ln(R/R0) versus ln A is a straight line passing through the origin with slope 1/3.

Answer:

(d) Fpp < Fpn = Fnn

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.

Fpp < Fpn = Fnn

Here, Fpp, Fpn and Fnn 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.

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Question 7:

(d) Fpp < Fpn = Fnn

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.

Fpp < Fpn = Fnn

Here, Fpp, Fpn and Fnn 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.

Answer:

(b) Fpp = Fpn = Fnn

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.

Fpp = Fpn = Fnn
Here, Fpp, Fpn and Fnn 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.

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Question 8:

(b) Fpp = Fpn = Fnn

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.

Fpp = Fpn = Fnn
Here, Fpp, Fpn and Fnn 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.

Answer:

(b) Fe >> Fn

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. Fe >> Fn



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Question 9:

(b) Fe >> Fn

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. Fe >> Fn

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.

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Question 10:

(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.

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. 

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Question 11:

(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. 

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 
τ=1λ=Τ120.693
Here, τ, λ and Τ12  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.

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Question 12:

(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 
τ=1λ=Τ120.693
Here, τ, λ and Τ12  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.

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.

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Question 13:

(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.

Answer:

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

Negative beta decay is given as
np+e-+ν¯  
Neutron decays to produce proton, electron and anti-neutrino.

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Question 14:

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

Negative beta decay is given as
np+e-+ν¯  
Neutron decays to produce proton, electron and anti-neutrino.

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 26 = 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.

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Question 15:

(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 26 = 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.

Answer:

(b) (ln 2/λ) and 1/λ

The half-life of a radioactive sample (t12) 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 N0.
No2=Noe-λt12t12=ln 2λ

Average life of the nuclei, tav=SNo=1λ
Here, S is the sum of all the lives of all the N nuclei that were active at t = 0 and λ is the decay constant of the sample.

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Question 16:

(b) (ln 2/λ) and 1/λ

The half-life of a radioactive sample (t12) 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 N0.
No2=Noe-λt12t12=ln 2λ

Average life of the nuclei, tav=SNo=1λ
Here, S is the sum of all the lives of all the N nuclei that were active at t = 0 and λ is the decay constant of the sample.

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,
H24e+N714O617+p11
So, the emitted particle is a proton.

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Question 17:

(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,
H24e+N714O617+p11
So, the emitted particle is a proton.

Answer:

(a) 10 g

57Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11×10-31 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.

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Question 18:

(a) 10 g

57Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11×10-31 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.

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.

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Question 19:

(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.

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.

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Question 1:

(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.

Answer:

(c) Density

Radius of a nucleus with mass number A is given as
 R=RoA13 
Here, Ro = 1.2 fm
∴ Volume of the nucleus = 4πR33=4πRo3A3 

This depends on A. With an increase in A, V increases proportionally.
Mass of the nucleus  AmN
Here, mN is the mass of a nucleon.
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 = MassVolume=AmN4πR33=AmN4πRo3A3=mN4πRo33=3mN4πRo3
This is independent of A and hence does not change as mass number increases.

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Question 2:

(c) Density

Radius of a nucleus with mass number A is given as
 R=RoA13 
Here, Ro = 1.2 fm
∴ Volume of the nucleus = 4πR33=4πRo3A3 

This depends on A. With an increase in A, V increases proportionally.
Mass of the nucleus  AmN
Here, mN is the mass of a nucleon.
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 = MassVolume=AmN4πR33=AmN4πRo3A3=mN4πRo33=3mN4πRo3
This is independent of A and hence does not change as mass number increases.

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).

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Question 3:

(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).

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.

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Question 4:

(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.

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.

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Question 5:

(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.

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.
 XZAYZ-2A-4+H24e
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. 
XZAYZ+1A+e+ν¯
 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.
XZAYZ-1A+β++ν
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.

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Question 6:

(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.
 XZAYZ-2A-4+H24e
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. 
XZAYZ+1A+e+ν¯
 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.
XZAYZ-1A+β++ν
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.

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.
 XZAYZ-2A-4+H24e
 During  β-decay​,  a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the  atomic number.
XZAYZ+1A+e-+ν¯
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. ​
XZAYZ-1A+β++ν
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:

(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.
 XZAYZ-2A-4+H24e
 During  β-decay​,  a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the  atomic number.
XZAYZ+1A+e-+ν¯
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. ​
XZAYZ-1A+β++ν
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.

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 F=q(v×B).
Here, 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:

(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 F=q(v×B).
Here, 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.

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:

(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.

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:

(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.

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:

(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.

Answer:

Given:
Mass of the nucleus, M = Amp
Volume of the nucleus, V = 43πR03A

Density of the matter, d=MV=Amp43πR03A
                                               =3mp4×πR03=3×1.0072764×3.14(1.1)3= 3×1017 kg/m3

Specific gravity of the nuclear matter = Density of matter Density of water
  Specific gravity = 3×1017103 = 3 × 1014 kg/m3

Page No 442:

Question 2:

Given:
Mass of the nucleus, M = Amp
Volume of the nucleus, V = 43πR03A

Density of the matter, d=MV=Amp43πR03A
                                               =3mp4×πR03=3×1.0072764×3.14(1.1)3= 3×1017 kg/m3

Specific gravity of the nuclear matter = Density of matter Density of water
  Specific gravity = 3×1017103 = 3 × 1014 kg/m3

Answer:

Given:
Mass of the neutron star, M = 4.0 × 1030 kg
Density of nucleus, d = 2.4×1017

Density of nucleus, d=MV

Here, V is the volume of the nucleus.

 V=Md=4×10302.4×1017        =10.6×1013        =16×1014

If R is the radius, then the volume of the neutron star is given by
V=43πR3  16×1014=43×π×R3 R3=16×34×1π×1014 R3=18×100π×1012 R=12×104×3.17       =1.585×104=15 km

Page No 442:

Question 3:

Given:
Mass of the neutron star, M = 4.0 × 1030 kg
Density of nucleus, d = 2.4×1017

Density of nucleus, d=MV

Here, V is the volume of the nucleus.

 V=Md=4×10302.4×1017        =10.6×1013        =16×1014

If R is the radius, then the volume of the neutron star is given by
V=43πR3  16×1014=43×π×R3 R3=16×34×1π×1014 R3=18×100π×1012 R=12×104×3.17       =1.585×104=15 km

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=Zmp+Nmn-Mc2

Here, mp = Mass of proton
mn = Mass of neutron
Z= Number of protons
N = Number of neutrons
c = Speed of light

On substituting the respective values, we have
28.2 =(2×1.007276+2×1.008665 -x)c2 x=4.0016 u

Page No 442:

Question 4:

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=Zmp+Nmn-Mc2

Here, mp = Mass of proton
mn = Mass of neutron
Z= Number of protons
N = Number of neutrons
c = Speed of light

On substituting the respective values, we have
28.2 =(2×1.007276+2×1.008665 -x)c2 x=4.0016 u

Answer:

Given:
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.
Reaction:
Li7+pα+α+E,
Energy release E is given by
E=mLi7+mp-2×mHe4c2=7.0160 u+1.007276 u-24.0026 uc2=(8.023273 u-8.0052 u) c2=0.018076 ×931 MeV =16.83 MeV

Page No 442:

Question 5:

Given:
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.
Reaction:
Li7+pα+α+E,
Energy release E is given by
E=mLi7+mp-2×mHe4c2=7.0160 u+1.007276 u-24.0026 uc2=(8.023273 u-8.0052 u) c2=0.018076 ×931 MeV =16.83 MeV

Answer:

Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, B=(Zmp+Nmn-M)c2                                                                
Here, mp = Mass of proton
M = Mass of nucleus
mn = Mass of neutron
c = Speed of light
On substituting the respective values, we get
B=[(79×1.007276+118×1.008665) u-196.96 u]c2  =198.597274-196.96×931 MeV  =1524.302094 MeV
Binding energy per nucleon =1524.3197=7.737 MeV

Page No 442:

Question 6:

Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, B=(Zmp+Nmn-M)c2                                                                
Here, mp = Mass of proton
M = Mass of nucleus
mn = Mass of neutron
c = Speed of light
On substituting the respective values, we get
B=[(79×1.007276+118×1.008665) u-196.96 u]c2  =198.597274-196.96×931 MeV  =1524.302094 MeV
Binding energy per nucleon =1524.3197=7.737 MeV

Answer:

(a)
Given:
Atomic mass of 238U, m(238U) = 238.0508 u
Atomic mass of 234Th, m(234Th)  = 234.04363 u
Atomic mass of 4He, m(4He) = 4.00260 u
When 238U emits an α-particle, the reaction is given by
 U238Th234+He4
Mass defect, Δm = [m(238U)-(m(234Th)+m(4He))]
Δm = [238.0508 - (234.04363 + 4.00260) = 0.00457 u
Energy released (E) when 238U emits an α-particle is given by
E=m c2E = [0.00457 u]×931.5 MeVE = 4.25467 MeV = 4.255 MeV

(b)

When two protons and two neutrons are emitted one by one, the reaction will be
U233Th234+2n+2p

Mass defect, m=mU238-[mTh234+2mn+2mp]m=238.0508 u-[234.04363 u+2(1.008665) u+2(1.007276) u]m= 0.024712 u

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

E=mc2E=0.024712 ×931.5 MeVE=23.019=23.02 MeV

Page No 442:

Question 7:

(a)
Given:
Atomic mass of 238U, m(238U) = 238.0508 u
Atomic mass of 234Th, m(234Th)  = 234.04363 u
Atomic mass of 4He, m(4He) = 4.00260 u
When 238U emits an α-particle, the reaction is given by
 U238Th234+He4
Mass defect, Δm = [m(238U)-(m(234Th)+m(4He))]
Δm = [238.0508 - (234.04363 + 4.00260) = 0.00457 u
Energy released (E) when 238U emits an α-particle is given by
E=m c2E = [0.00457 u]×931.5 MeVE = 4.25467 MeV = 4.255 MeV

(b)

When two protons and two neutrons are emitted one by one, the reaction will be
U233Th234+2n+2p

Mass defect, m=mU238-[mTh234+2mn+2mp]m=238.0508 u-[234.04363 u+2(1.008665) u+2(1.007276) u]m= 0.024712 u

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

E=mc2E=0.024712 ×931.5 MeVE=23.019=23.02 MeV

Answer:

Given:
Atomic mass of 223Ra, m(223Ra) = 223.018 u
Atomic mass of 209Pb, m(209Pb) = 208.981 u 
Atomic mass of 14C, m(14C) = 14.003 u

Reaction: 

Ra223209Pb+C14

Energy, E=mR223a-mP209b+mC14c2                  = 223.018 u-208.981+14.003 u c2                   =0.034×931 MeV                   =31.65 MeV

Page No 442:

Question 8:

Given:
Atomic mass of 223Ra, m(223Ra) = 223.018 u
Atomic mass of 209Pb, m(209Pb) = 208.981 u 
Atomic mass of 14C, m(14C) = 14.003 u

Reaction: 

Ra223209Pb+C14

Energy, E=mR223a-mP209b+mC14c2                  = 223.018 u-208.981+14.003 u c2                   =0.034×931 MeV                   =31.65 MeV

Answer:

Given:
Mass of an atom with Z protons and N neutrons = MZ,N
Mass of hydrogen atom = MH
As hydrogen contains only protons, the reaction will be given by
EZ,NEZ-1,N+p1EZ,NEz-1,N+1H1
∴ Minimum energy needed to separate a proton, E=(MZ-1,N+MH-MZ,N)c2

Page No 442:

Question 9:

Given:
Mass of an atom with Z protons and N neutrons = MZ,N
Mass of hydrogen atom = MH
As hydrogen contains only protons, the reaction will be given by
EZ,NEZ-1,N+p1EZ,NEz-1,N+1H1
∴ Minimum energy needed to separate a proton, E=(MZ-1,N+MH-MZ,N)c2

Answer:

Before the separation of neutron, let the mass of the nucleus be MZ,N.
Then, after the separation of neutron, the mass of the nucleus will be MZ,N-1.
The reaction is given by
EZ,N=EZ,N-1+n01
If MN is the mass of the neutron, then the energy needed to separate the neutron bE will be
E = (Final mass of nucleus + Mass of neutron − Initial mass of the nucleus)c2
E=(MZ,N-1+MN-MZ,N)c2

Page No 442:

Question 10:

Before the separation of neutron, let the mass of the nucleus be MZ,N.
Then, after the separation of neutron, the mass of the nucleus will be MZ,N-1.
The reaction is given by
EZ,N=EZ,N-1+n01
If MN is the mass of the neutron, then the energy needed to separate the neutron bE will be
E = (Final mass of nucleus + Mass of neutron − Initial mass of the nucleus)c2
E=(MZ,N-1+MN-MZ,N)c2

Answer:

Given:
Atomic mass of 32P, m(32P) = 31.974 u
Atomic mass of 32S, m(32S) = 31.972 u
Reaction:
P32S32+1v0+-1β0

Energy of antineutrino and β-particle, E = [m(32P)-m(32S)]c2
                                                         = (31.974 u− 31.972 u)c2
                                                         = 0.002 × 931 = 1.862 MeV

Page No 442:

Question 11:

Given:
Atomic mass of 32P, m(32P) = 31.974 u
Atomic mass of 32S, m(32S) = 31.972 u
Reaction:
P32S32+1v0+-1β0

Energy of antineutrino and β-particle, E = [m(32P)-m(32S)]c2
                                                         = (31.974 u− 31.972 u)c2
                                                         = 0.002 × 931 = 1.862 MeV

Answer:

Given:
Half-life period of free neutron beta-decays to a proton, T1/2 = 14 minutes
 
Half-life period, T1/2  = 0.6931λ

Here, λ = Decay constant

 λ=0.69314×60      =8.25×10-4 s-1

If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.

 ∴ Energy liberated, E=[mn-mp+me]c2                                       =[1.008665 u-1.007276+0.0005486 u]c2                                    =0.0008404×931 MeV                                    =782 keV

Page No 442:

Question 12:

Given:
Half-life period of free neutron beta-decays to a proton, T1/2 = 14 minutes
 
Half-life period, T1/2  = 0.6931λ

Here, λ = Decay constant

 λ=0.69314×60      =8.25×10-4 s-1

If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.

 ∴ Energy liberated, E=[mn-mp+me]c2                                       =[1.008665 u-1.007276+0.0005486 u]c2                                    =0.0008404×931 MeV                                    =782 keV

Answer:

(a) Ra88226α24+Rn86222(b) O819F919+e¯+v¯(c) Ar1325Mg1225+e++v

Page No 442:

Question 13:

(a) Ra88226α24+Rn86222(b) O819F919+e¯+v¯(c) Ar1325Mg1225+e++v

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=Ec
 Here, E = Energy of neutrino
  c = Speed of light
 
  P=500×1.6×10-193×108×103         =2.67×10-22 Kgms-1

Page No 442:

Question 14:

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=Ec
 Here, E = Energy of neutrino
  c = Speed of light
 
  P=500×1.6×10-193×108×103         =2.67×10-22 Kgms-1

Answer:

(a) Decay of potassium-40 by βemission is given by
     K401920Ca40+β-+v¯
    Decay of potassium-40 by β+ emission is given by
    K401918Ar40+β++v
    Decay of potassium-40 by electron capture is given by
    K4019+e-18Ar40+v

(b) 
  Qvalue in the β decay is given by
  Qvalue = [m(19K40) − m(20Ca40)]c2
            = [39.9640 u − 39.9626 u]c2
           = 0.0014 × 931 MeV
           = 1.3034 MeV
Qvalue in the β+ decay is given by
 Qvalue = [m(19K40) − m(20Ar40) − 2me]c2
           = [39.9640 u − 39.9624 u −  0.0021944 u]c2
           = (39.9640 − 39.9624) 931 MeV − 1022 keV
           = 1489.96 keV − 1022 keV
           = 0.4679 MeV  
Qvalue in the electron capture is given by
Qvalue = [ m(19K40) − m(20Ar40)]c2
          = (39.9640 − 39.9624)uc2
          = 1.4890 = 1.49 MeV

Page No 442:

Question 15:

(a) Decay of potassium-40 by βemission is given by
     K401920Ca40+β-+v¯
    Decay of potassium-40 by β+ emission is given by
    K401918Ar40+β++v
    Decay of potassium-40 by electron capture is given by
    K4019+e-18Ar40+v

(b) 
  Qvalue in the β decay is given by
  Qvalue = [m(19K40) − m(20Ca40)]c2
            = [39.9640 u − 39.9626 u]c2
           = 0.0014 × 931 MeV
           = 1.3034 MeV
Qvalue in the β+ decay is given by
 Qvalue = [m(19K40) − m(20Ar40) − 2me]c2
           = [39.9640 u − 39.9624 u −  0.0021944 u]c2
           = (39.9640 − 39.9624) 931 MeV − 1022 keV
           = 1489.96 keV − 1022 keV
           = 0.4679 MeV  
Qvalue in the electron capture is given by
Qvalue = [ m(19K40) − m(20Ar40)]c2
          = (39.9640 − 39.9624)uc2
          = 1.4890 = 1.49 MeV

Answer:

The nuclear process taking place is shown below.
Li86+n Li37Li37+nLi38Be48+v¯+e-Be48He24+He24

Page No 442:

Question 16:

The nuclear process taking place is shown below.
Li86+n Li37Li37+nLi38Be48+v¯+e-Be48He24+He24

Answer:

Given:
Mass of 11C, m(11C) = 11.0114 u
Mass of 11B, m(11B) = 11.0093 u
Energy liberated in the β+ decay (Q) is given by
Q=mC11-mB11-2mec2
   = (11.0114 u − 11.0093 u - 2×0.0005486 u)c2
   = 0.0010028 × 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:

Given:
Mass of 11C, m(11C) = 11.0114 u
Mass of 11B, m(11B) = 11.0093 u
Energy liberated in the β+ decay (Q) is given by
Q=mC11-mB11-2mec2
   = (11.0114 u − 11.0093 u - 2×0.0005486 u)c2
   = 0.0010028 × 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

Answer:

Given:
Atomic mass of 228Th, m(228Th) = 228.028726 u
Atomic mass of 224Ra, m(224Ra) = 224.020196 u
Atomic mass of H24, m(H24) = 4.00260 u
Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, K = mTh228-mRa224+mH24c2
                                                        = (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
                                                        = 5.30383 MeV = 5.304 MeV

Page No 442:

Question 18:

Given:
Atomic mass of 228Th, m(228Th) = 228.028726 u
Atomic mass of 224Ra, m(224Ra) = 224.020196 u
Atomic mass of H24, m(H24) = 4.00260 u
Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, K = mTh228-mRa224+mH24c2
                                                        = (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
                                                        = 5.30383 MeV = 5.304 MeV

Answer:

 Given:
 Atomic mass of 12N, m(12N) = 12.018613 u
 12N → 12C* + e+ + v
 12C* → 12C + γ (4.43 MeV)

 Net reaction is given by
 12N → 12C + e+ + v + γ (4.43 MeV)
 
Qvalue  of the β+ decay will be
 Qvalue= [ m(12N) - (m(12C*) + 2me)]c2
           = [12.018613 ×931 MeV - (12×931 + 4.43) MeV - (2×511) 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:

 Given:
 Atomic mass of 12N, m(12N) = 12.018613 u
 12N → 12C* + e+ + v
 12C* → 12C + γ (4.43 MeV)

 Net reaction is given by
 12N → 12C + e+ + v + γ (4.43 MeV)
 
Qvalue  of the β+ decay will be
 Qvalue= [ m(12N) - (m(12C*) + 2me)]c2
           = [12.018613 ×931 MeV - (12×931 + 4.43) MeV - (2×511) 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.

Answer:

Given:
Decay constant of Hg80197, λ = 1.8 × 10−4 s-1

(a)
Half-life, T12=0.693λ

 T1/2=0.6931.8×10-4            = 3850 s=64 minutes      

(b)     
Average life, Tav=T1/20.693                              =640.693                              =92 minutes

(c)
Number of active nuclei of mercury at t = 0
                                                                     = N0
                                                                    = 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N =  75

Now, NN0=e-λt
Here, N = Number of inactive nuclei
            N0 = Number of nuclei at t = 0
            λ = Disintegration constant

On substituting the values, we get
75100=e-λt
0.75=e-λx In 0.75=-λt t=In 0.75-0.00018       =1600 s



Page No 443:

Question 20:

Given:
Decay constant of Hg80197, λ = 1.8 × 10−4 s-1

(a)
Half-life, T12=0.693λ

 T1/2=0.6931.8×10-4            = 3850 s=64 minutes      

(b)     
Average life, Tav=T1/20.693                              =640.693                              =92 minutes

(c)
Number of active nuclei of mercury at t = 0
                                                                     = N0
                                                                    = 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N =  75

Now, NN0=e-λt
Here, N = Number of inactive nuclei
            N0 = Number of nuclei at t = 0
            λ = Disintegration constant

On substituting the values, we get
75100=e-λt
0.75=e-λx In 0.75=-λt t=In 0.75-0.00018       =1600 s

Answer:

Given:
Half-life of 199Au, T1/2= 2.7 days

Disintegration constant, λ=0.693T1/20.6932.7×24×60×60=2.97×10-6  s-1

Number of atoms left undecayed, N = 1×10-6×6.023×1023198

Now, activity, A0=λN
  
= 1×10-6×6.023×1023198×2.97×10-6
   = 0.244 Ci

(b) After 7 days,
Activity, A=A0e-λt  
Here, A0 = 0.244 Ci
 A =0.244×e-2.97×10-6×7×3600×24        =0.244×e17962.56×10-4       =0.040 Ci

Page No 443:

Question 21:

Given:
Half-life of 199Au, T1/2= 2.7 days

Disintegration constant, λ=0.693T1/20.6932.7×24×60×60=2.97×10-6  s-1

Number of atoms left undecayed, N = 1×10-6×6.023×1023198

Now, activity, A0=λN
  
= 1×10-6×6.023×1023198×2.97×10-6
   = 0.244 Ci

(b) After 7 days,
Activity, A=A0e-λt  
Here, A0 = 0.244 Ci
 A =0.244×e-2.97×10-6×7×3600×24        =0.244×e17962.56×10-4       =0.040 Ci

Answer:

Given:
Half-life of radioactive 131I, T1/2 = 8 days
Activity of the sample at t = 0, A0 = 20 µCi
Time, t = 4 days

(a) Disintegration constant, λ=0.693T1/2=0.0866
Activity (A) at t = 4 days is given by
 A=A0e-λtA=20×10-6×e-0.0866×4      =1.41×10-6 Ci=14 μCi

(b) Decay constant is a constant for a radioactive sample and depends only on its half-life.
  λ=0.693[8×24×3600]    =1.0026×10-6 s-1

Page No 443:

Question 22:

Given:
Half-life of radioactive 131I, T1/2 = 8 days
Activity of the sample at t = 0, A0 = 20 µCi
Time, t = 4 days

(a) Disintegration constant, λ=0.693T1/2=0.0866
Activity (A) at t = 4 days is given by
 A=A0e-λtA=20×10-6×e-0.0866×4      =1.41×10-6 Ci=14 μCi

(b) Decay constant is a constant for a radioactive sample and depends only on its half-life.
  λ=0.693[8×24×3600]    =1.0026×10-6 s-1

Answer:

Given:
Decay constant, λ = 4.9 × 10−18 s−1

(a) Average life of uranium τ is given by
  τ=1λ  =14.9×10-18  =14.9×1018 s  =10164.9×365×24×36 years =10164.9×365×24×36  years = 6.47×10-7×1016 years =6.47×109 years

(b) Half-life of uranium T12 is given by
T12=0.693λ=0.6934.9×10-18    =0.6934.9×1018 s    =0.1414×1018 s    =0.1414×1018365×24×3600    =1414×1012365×24×36   = 4.48×10-3×1012   =4.5×109 years

(c) Time, t = 9 × 109 years
 Activity (A) of the sample, at any time t, is given by
    A=A02tT1/2
  Here, A0 = Activity of the sample at t = 0
  A0A=29×1094.5×109=22=4

Page No 443:

Question 23:

Given:
Decay constant, λ = 4.9 × 10−18 s−1

(a) Average life of uranium τ is given by
  τ=1λ  =14.9×10-18  =14.9×1018 s  =10164.9×365×24×36 years =10164.9×365×24×36  years = 6.47×10-7×1016 years =6.47×109 years

(b) Half-life of uranium T12 is given by
T12=0.693λ=0.6934.9×10-18    =0.6934.9×1018 s    =0.1414×1018 s    =0.1414×1018365×24×3600    =1414×1012365×24×36   = 4.48×10-3×1012   =4.5×109 years

(c) Time, t = 9 × 109 years
 Activity (A) of the sample, at any time t, is given by
    A=A02tT1/2
  Here, A0 = Activity of the sample at t = 0
  A0A=29×1094.5×109=22=4

Answer:

Given:
Initial rate of decay, A0 = 500,
Rate of decay after 50 minutes, A = 200
Time, t = 50 min
            = 50 × 60
            = 3000 s

(a)
Activity, A = A0eλt
Here, λ = Disintegration constant
 200 = 500 × e−50×60×λ
25=e-3000λ In25=-3000λ λ=3.05×10-4 s-1

(b)
Half-life, T1/2=0.693λ                         =2272.13 s                         =38 min

Page No 443:

Question 24:

Given:
Initial rate of decay, A0 = 500,
Rate of decay after 50 minutes, A = 200
Time, t = 50 min
            = 50 × 60
            = 3000 s

(a)
Activity, A = A0eλt
Here, λ = Disintegration constant
 200 = 500 × e−50×60×λ
25=e-3000λ In25=-3000λ λ=3.05×10-4 s-1

(b)
Half-life, T1/2=0.693λ                         =2272.13 s                         =38 min

Answer:

Given:
Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 106 disintegration/sec
Time, t = 20 hours
Activity of radioactive sample A is given by
A=A02tT2Here, T1/2=Half-life periodOn substituting the values of A0 and A, we have 2tT2=22tT2=2T2=t/2=20 h/2=10 h

100 hours after the beginning,
Count rate, A"=A02tT2A"=4×1062100/10           =0.390625×104           =3.9×103 disintegrations/sec
     

Page No 443:

Question 25:

Given:
Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 106 disintegration/sec
Time, t = 20 hours
Activity of radioactive sample A is given by
A=A02tT2Here, T1/2=Half-life periodOn substituting the values of A0 and A, we have 2tT2=22tT2=2T2=t/2=20 h/2=10 h

100 hours after the beginning,
Count rate, A"=A02tT2A"=4×1062100/10           =0.390625×104           =3.9×103 disintegrations/sec
     

Answer:

Given:
Half-life of radium, T1/2 = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl2 = 226 + 71 = 297 g
297 g = 1 mole of RaCl2
0.1 g = 1297×0.1 mole of RaCl2
Total number of atoms in 0.1 g of RaCl2, N =0.1×6.023×1023297 = 0.02027×1022
∴ No of atoms, N = 0.02027 × 1022

  Disintegration constant, λ=0.693T12                                             =0.6931602×365×24×3600                                             =1.371×10-11

Activity of radioactive sample, A = λN
                                                      = 1.371×10-11×2.027×1020 
                                                      = 2.8×109 disintegrations/second

Page No 443:

Question 26:

Given:
Half-life of radium, T1/2 = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl2 = 226 + 71 = 297 g
297 g = 1 mole of RaCl2
0.1 g = 1297×0.1 mole of RaCl2
Total number of atoms in 0.1 g of RaCl2, N =0.1×6.023×1023297 = 0.02027×1022
∴ No of atoms, N = 0.02027 × 1022

  Disintegration constant, λ=0.693T12                                             =0.6931602×365×24×3600                                             =1.371×10-11

Activity of radioactive sample, A = λN
                                                      = 1.371×10-11×2.027×1020 
                                                      = 2.8×109 disintegrations/second

Answer:

Given:
Half-life of radioisotope, T1/2 = 10 hrs
Initial activity, A0 = 1 Ci
Disintegration constant, λ=0.69310×3600 s-1
Activity of radioactive sample, A=A0e-λt 
Here, A0 = Initial activity
λ = Disintegration constant
t = Time taken

After 9 hours,
Activity, A=A0e-λt=1×e-0.69310×3600×9=0.536 Ci 

∴ Number of atoms left, N = Aλ = 0.536×10×3.7×1010×36000.693=103.023×1013
A
fter 10 hrs, 

Activity, A"=A0e-λt    =1×e-0.69310×10=0.5 Ci

Number of atoms left after the 10th hour N" will be
A"=λN"N"=A"λ    =0.5×3.7×1010×3.6000.693/10    =26.37×1010×3600=96.103×1013

Number of disintegrations  = (103.023 − 96.103) × 1013
                                       = 6.92 × 1013

Page No 443:

Question 27:

Given:
Half-life of radioisotope, T1/2 = 10 hrs
Initial activity, A0 = 1 Ci
Disintegration constant, λ=0.69310×3600 s-1
Activity of radioactive sample, A=A0e-λt 
Here, A0 = Initial activity
λ = Disintegration constant
t = Time taken

After 9 hours,
Activity, A=A0e-λt=1×e-0.69310×3600×9=0.536 Ci 

∴ Number of atoms left, N = Aλ = 0.536×10×3.7×1010×36000.693=103.023×1013
A
fter 10 hrs, 

Activity, A"=A0e-λt    =1×e-0.69310×10=0.5 Ci

Number of atoms left after the 10th hour N" will be
A"=λN"N"=A"λ    =0.5×3.7×1010×3.6000.693/10    =26.37×1010×3600=96.103×1013

Number of disintegrations  = (103.023 − 96.103) × 1013
                                       = 6.92 × 1013

Answer:

Given:
Half-life of 32P source, T12= 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, A0 = 800 disintegration/sec
Disintegration constant λ is given by
λ=0.693T12 = 0.69314.3 days-1
Activity A is given by
A = A0eλt
Here, λ = Disintegration constant   
Activity of the radioactive isotope after one month (selling rate of the radioactive isotope) A is given below.     
 A=800×e-0.69314.3×30       =800×0.233669      =186.935=Rs 187 

Page No 443:

Question 28:

Given:
Half-life of 32P source, T12= 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, A0 = 800 disintegration/sec
Disintegration constant λ is given by
λ=0.693T12 = 0.69314.3 days-1
Activity A is given by
A = A0eλt
Here, λ = Disintegration constant   
Activity of the radioactive isotope after one month (selling rate of the radioactive isotope) A is given below.     
 A=800×e-0.69314.3×30       =800×0.233669      =186.935=Rs 187 

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:

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.

Answer:

(a) The reaction is given by
    C6B5+β++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,  λ=0.693T12 = 0.69320.3 min-1

If t is the time taken by the mixture in converting, let the total no. of atoms be 100N0.

  Carbon Boron
Initial 90 N0 10 N0
Final 10 N0 90 N0
  N = N0eλt
Here, N0 = Initial number of atoms
N = Number of atoms left undecayed
 10N0 =90N0eλt  ( For carbon)
 19=e-0.69320.3×t   In19=-0.69320.3t t= 64.36=64 min

Page No 443:

Question 30:

(a) The reaction is given by
    C6B5+β++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,  λ=0.693T12 = 0.69320.3 min-1

If t is the time taken by the mixture in converting, let the total no. of atoms be 100N0.

  Carbon Boron
Initial 90 N0 10 N0
Final 10 N0 90 N0
  N = N0eλt
Here, N0 = Initial number of atoms
N = Number of atoms left undecayed
 10N0 =90N0eλt  ( For carbon)
 19=e-0.69320.3×t   In19=-0.69320.3t t= 64.36=64 min

Answer:

Given:
Number of tritium atoms, N0 = 4 × 1023
Half-life of tritium nuclei, T12= 12.3 years

Disintegration constant,  λ=0.693T12=0.69312.3 years-1

Activity of the sample A is given by 
A0 = dNdt = λN0
A0=0.693t1/2N0      =0.69312.3×4×1023 disintegration/year     =0.693×4×102312.3×3600×24×365 disintegration/sec     =7.146×1014 disintegration/sec

(b) Activity of the sample, A = 7.146×1014 disintegration/sec

   Number of decays in the next 10 hours=7.146×1014×10×3600                                                                     =257.256×1017                                                                       =2.57×1019

(c) Number of atoms left undecayed, N=N0e-λt
   Here, N0 = Initial number of atoms
    N=4×1023×e-0.69312.3×6.15=2.83×1023
 
Number of atoms disintegrated = (N0-N) = (4-2.83) × 1023 = 1.17 × 1023

Page No 443:

Question 31:

Given:
Number of tritium atoms, N0 = 4 × 1023
Half-life of tritium nuclei, T12= 12.3 years

Disintegration constant,  λ=0.693T12=0.69312.3 years-1

Activity of the sample A is given by 
A0 = dNdt = λN0
A0=0.693t1/2N0      =0.69312.3×4×1023 disintegration/year     =0.693×4×102312.3×3600×24×365 disintegration/sec     =7.146×1014 disintegration/sec

(b) Activity of the sample, A = 7.146×1014 disintegration/sec

   Number of decays in the next 10 hours=7.146×1014×10×3600                                                                     =257.256×1017                                                                       =2.57×1019

(c) Number of atoms left undecayed, N=N0e-λt
   Here, N0 = Initial number of atoms
    N=4×1023×e-0.69312.3×6.15=2.83×1023
 
Number of atoms disintegrated = (N0-N) = (4-2.83) × 1023 = 1.17 × 1023

Answer:

Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 1016
Total counts radiated from the source, dNdt = Total surface area × 50000 counts/cm2
                                                                     = 4 × 3.14 × 1 × 104 × 5 × 104
                                                                     = 6.28 × 109 Counts
We know
dNdt=λN

Here, λ = Disintegration constant
  λ=6.28×1096×1016       =1.0467×10-7       =1.05×10-7s-1

Page No 443:

Question 32:

Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 1016
Total counts radiated from the source, dNdt = Total surface area × 50000 counts/cm2
                                                                     = 4 × 3.14 × 1 × 104 × 5 × 104
                                                                     = 6.28 × 109 Counts
We know
dNdt=λN

Here, λ = Disintegration constant
  λ=6.28×1096×1016       =1.0467×10-7       =1.05×10-7s-1

Answer:

Given:
Half-life of 238U, t1/2 = 4.47 × 109 years
Total number of atoms present in the rock initially, N0=6.023×1023×2238+6.023×1023×0.6206                                                                                                =12.046238+3.62206×1020                                                                                                =0.0505+0.0175×1020                                                                                                =0.0680×1020

Now, N = N0e-λt
Here, λ = Disintegration constant
t = Life of the rock
 N=N0e-0.693t1/2×t 0.0505=0.0680e-0.6934.47×109×t ln0.05050.0680=-0.69314.47×109×tt=1.92×109 years

Page No 443:

Question 33:

Given:
Half-life of 238U, t1/2 = 4.47 × 109 years
Total number of atoms present in the rock initially, N0=6.023×1023×2238+6.023×1023×0.6206                                                                                                =12.046238+3.62206×1020                                                                                                =0.0505+0.0175×1020                                                                                                =0.0680×1020

Now, N = N0e-λt
Here, λ = Disintegration constant
t = Life of the rock
 N=N0e-0.693t1/2×t 0.0505=0.0680e-0.6934.47×109×t ln0.05050.0680=-0.69314.47×109×tt=1.92×109 years

Answer:

Given:
Initial activity of charcoal, A0 = 15.3 disintegrations per gram per minute
Half-life of charcoal, T12= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute

Disintegration constant, λ=0.693T12 = 0.6935370  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=A0e-λt
A=A0e-0.6935730×tIn12.315.3=-0.6935730t0.218253=0.6935730×tt=1804.3 years

Page No 443:

Question 34:

Given:
Initial activity of charcoal, A0 = 15.3 disintegrations per gram per minute
Half-life of charcoal, T12= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute

Disintegration constant, λ=0.693T12 = 0.6935370  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=A0e-λt
A=A0e-0.6935730×tIn12.315.3=-0.6935730t0.218253=0.6935730×tt=1804.3 years

Answer:

Given:
Half-life time of tritium, T12 = 12.5 years
Disintegration constant, λ=0.69312.5  per year
Let A0 be the activity, when the bottle was manufactured.
Activity after 8 years A is given by
A=A0e-0.69312.5×8     ...(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle A' on the mountain is given by
A'=A0e-λt        
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
A'=A0e-0.69312.5×8×0.015     ...(2)
Comparing (1) and (2)
0.69312.5 t=-0.6931×812.5+In [0.015] -0.69312.5 t=-0.69312.5×8-4.19970.693 t=58.040 t=83.75 years

Page No 443:

Question 35:

Given:
Half-life time of tritium, T12 = 12.5 years
Disintegration constant, λ=0.69312.5  per year
Let A0 be the activity, when the bottle was manufactured.
Activity after 8 years A is given by
A=A0e-0.69312.5×8     ...(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle A' on the mountain is given by
A'=A0e-λt        
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
A'=A0e-0.69312.5×8×0.015     ...(2)
Comparing (1) and (2)
0.69312.5 t=-0.6931×812.5+In [0.015] -0.69312.5 t=-0.69312.5×8-4.19970.693 t=58.040 t=83.75 years

Answer:

(a) For t = 0,
lnR0R=In30×10930×109=0
For t = 25 s,
lnR0R2=In30×10916×109=0.63
For t = 50 s,
InR0R3=In30×1098×109=1.35
For t = 75 s,
lnR0R4=In30×1093.8×109=2.06
For t = 100 s,
InR0R5=In30×1092×109=2.7
The required graph is shown below.


(b) Slope of the graph = 0.028
∴ Decay constant, λ = 0.028 min-1
The half-life periodT12 is given by
    T12=0.693λ     =0.6930.028=25 min

Page No 443:

Question 36:

(a) For t = 0,
lnR0R=In30×10930×109=0
For t = 25 s,
lnR0R2=In30×10916×109=0.63
For t = 50 s,
InR0R3=In30×1098×109=1.35
For t = 75 s,
lnR0R4=In30×1093.8×109=2.06
For t = 100 s,
InR0R5=In30×1092×109=2.7
The required graph is shown below.


(b) Slope of the graph = 0.028
∴ Decay constant, λ = 0.028 min-1
The half-life periodT12 is given by
    T12=0.693λ     =0.6930.028=25 min

Answer:

Given:
Half-life period of 40K, T12 = 1.30 × 109 years
Count given by 1 g of pure KCI, A = 160 counts/s

Disintegration constant, λ=0.693T12
Now, activity, A = λN

160=0.693t1/2 ×N 160=0.6931.30×109×365×86400× N N=160×1.30×365×86400×1090.693N=9.5×1018   
6.023 × 1023 atoms are present in 40 gm.

Thus, 9.5 × 1018 atoms will be present in40×9.5×10186.023×1023 gm.=4×9.5×10-46.023 gm=6.309×10-4=0.00063 gm
Relative abundance of 40K  in natural potassium = (2 × 0.00063 × 100)% = 0.12%

Page No 443:

Question 37:

Given:
Half-life period of 40K, T12 = 1.30 × 109 years
Count given by 1 g of pure KCI, A = 160 counts/s

Disintegration constant, λ=0.693T12
Now, activity, A = λN

160=0.693t1/2 ×N 160=0.6931.30×109×365×86400× N N=160×1.30×365×86400×1090.693N=9.5×1018   
6.023 × 1023 atoms are present in 40 gm.

Thus, 9.5 × 1018 atoms will be present in40×9.5×10186.023×1023 gm.=4×9.5×10-46.023 gm=6.309×10-4=0.00063 gm
Relative abundance of 40K  in natural potassium = (2 × 0.00063 × 100)% = 0.12%

Answer:

Given:
Decay constant of electron capture = 0.257 per day

(a) The reaction is given as
 Hg80197+eAu79197+v
The other particle emitted in this reaction is neutrino v.

(b) Moseley's law is given by
v = a(Zb)

We  know
v=cλ
Here, c = Speed of light
        λ = Wavelength of the Kα X-ray

cλ=4.95×107(79-1)         =4.95×107×78cλ=(4.95×78)2×1014λ=3×108149073.21×1014   =20 pm

Page No 443:

Question 38:

Given:
Decay constant of electron capture = 0.257 per day

(a) The reaction is given as
 Hg80197+eAu79197+v
The other particle emitted in this reaction is neutrino v.

(b) Moseley's law is given by
v = a(Zb)

We  know
v=cλ
Here, c = Speed of light
        λ = Wavelength of the Kα X-ray

cλ=4.95×107(79-1)         =4.95×107×78cλ=(4.95×78)2×1014λ=3×108149073.21×1014   =20 pm

Answer:

Given:
Half life period of isotope = t1/2

Disintegration constant, λ=0.693t1/2

Rate of Radio active decay R is given by,
  R=dNdt
We are to show that after time t >> t1/2 the number of active nuclei is constant.
dNdtPresent=R=dNdtdecay R = dNdtdecay
Rate of radioactive decay, R=λN
Here, λ = Radioactive decay constant
N = Constant number
R=0.693t1/2×NRt1/2=0.693NN=Rt1/20.693
This value of N should be constant.

Page No 443:

Question 39:

Given:
Half life period of isotope = t1/2

Disintegration constant, λ=0.693t1/2

Rate of Radio active decay R is given by,
  R=dNdt
We are to show that after time t >> t1/2 the number of active nuclei is constant.
dNdtPresent=R=dNdtdecay R = dNdtdecay
Rate of radioactive decay, R=λN
Here, λ = Radioactive decay constant
N = Constant number
R=0.693t1/2×NRt1/2=0.693NN=Rt1/20.693
This value of N should be constant.

Answer:

Let the number of atoms present at t = 0 be N0.
Let N be the number of radio-active isotopes present at time t.
Then,
N = N0e−λt
Here, λ = Disintegration constant
∴ Number of radioactive isotopes decayed = N0N = N0N0eλt
                                                                     = N0 (1−eλt)   ...(1)
Rate of decay R is given by
 R = λN0   ...(2)
Substituting the value of N0 from equation (2) to equation (1), we get
N=N0(1-e-λt)  =Rλ(1-e-λt)

Page No 443:

Question 40:

Let the number of atoms present at t = 0 be N0.
Let N be the number of radio-active isotopes present at time t.
Then,
N = N0e−λt
Here, λ = Disintegration constant
∴ Number of radioactive isotopes decayed = N0N = N0N0eλt
                                                                     = N0 (1−eλt)   ...(1)
Rate of decay R is given by
 R = λN0   ...(2)
Substituting the value of N0 from equation (2) to equation (1), we get
N=N0(1-e-λt)  =Rλ(1-e-λt)

Answer:

Given:
Initial no of atoms, N0 = 1 mole = 6 × 1023 atoms
Half-life of the radioactive material, T1/2 = 14.3 days
Time taken by the plant to settle down, t = 70 h

Disintegration constant, λ=0.693t1/2 = 0.69314.3×24 h-1
N = N0e−λt
   =6×1023×e-0.693×7014.3×24=6×1023×0.868=5.209×1023

 Activity, R=dNdt=5.209×1023×0.69314.3×24  =0.0105×10233600dis/hr =2.9×10-6×1023 dis/sec =2.9×1017dis/sec

Fraction of activity transmitted = 1 μCi2.9×1017×100%                                                     =1×3.7×1042.9×1017×100%                                                     =1.275×10-11%

Page No 443:

Question 41:

Given:
Initial no of atoms, N0 = 1 mole = 6 × 1023 atoms
Half-life of the radioactive material, T1/2 = 14.3 days
Time taken by the plant to settle down, t = 70 h

Disintegration constant, λ=0.693t1/2 = 0.69314.3×24 h-1
N = N0e−λt
   =6×1023×e-0.693×7014.3×24=6×1023×0.868=5.209×1023

 Activity, R=dNdt=5.209×1023×0.69314.3×24  =0.0105×10233600dis/hr =2.9×10-6×1023 dis/sec =2.9×1017dis/sec

Fraction of activity transmitted = 1 μCi2.9×1017×100%                                                     =1×3.7×1042.9×1017×100%                                                     =1.275×10-11%

Answer:

Given:
Volume of the vessel, V = 125 cm3 = 0.125 L
Half-life time of tritium, t1/2 = 12.3 y = 3.82 × 108 s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,

Disintegration constant,  λ=0.693t1/2
                                      =0.6933.82×108=0.1814×10-8=1.81×10-9 s-1

No. of atoms left undecayed, N = n × 6.023 × 1023
                                             =5×0.1258.2×10-2×3×102×6.023×1023    n=PVRT=1.5×1022 atoms

Activity, A = λN
       = 1.81 × 10−9 × 1.5 × 1022 = 2.7 × 1013 disintegration/sec
   
A=2.7×10133.7×1010 Ci=729.72 Ci



Page No 444:

Question 42:

Given:
Volume of the vessel, V = 125 cm3 = 0.125 L
Half-life time of tritium, t1/2 = 12.3 y = 3.82 × 108 s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,

Disintegration constant,  λ=0.693t1/2
                                      =0.6933.82×108=0.1814×10-8=1.81×10-9 s-1

No. of atoms left undecayed, N = n × 6.023 × 1023
                                             =5×0.1258.2×10-2×3×102×6.023×1023    n=PVRT=1.5×1022 atoms

Activity, A = λN
       = 1.81 × 10−9 × 1.5 × 1022 = 2.7 × 1013 disintegration/sec
   
A=2.7×10133.7×1010 Ci=729.72 Ci

Answer:

 Given:
Half-life of 212Bi, T1/2 = 1 h-1
When Bi83212 disintegrates by emitting an α-particle

Bi83212T812081+He24(α)

When Bi83212 disintegrates by emitting a βparticle

Bi83212P084212+β-+v¯

Half-life period of 212Bi, T12= 1 h-1
At t = 0, the amount of 212Bi present = 1 g

 At t = 1 = One half-life,
Amount of  212Bi present = 0.5 g
Probability of disintegration of α-decay and β-decay are in the ratio 713.
In 20 g of 212Bi, the amount of 208Ti formed = 7 g
In 1 g of 212Bi, the amount of 208Ti formed = 7/20 g
∴ Amount of 208Ti present in 0.5 g = 720×0.5=0.175 g

In 20 g of 212Bi, the amount of 212Po formed = 13 g
In 1 g of 212Bi, the amount of 212Po formed = 13/20 g
∴ Amount of 212Po present in 0.5 g = 1320×0.5=0.325 g

Page No 444:

Question 43:

 Given:
Half-life of 212Bi, T1/2 = 1 h-1
When Bi83212 disintegrates by emitting an α-particle

Bi83212T812081+He24(α)

When Bi83212 disintegrates by emitting a βparticle

Bi83212P084212+β-+v¯

Half-life period of 212Bi, T12= 1 h-1
At t = 0, the amount of 212Bi present = 1 g

 At t = 1 = One half-life,
Amount of  212Bi present = 0.5 g
Probability of disintegration of α-decay and β-decay are in the ratio 713.
In 20 g of 212Bi, the amount of 208Ti formed = 7 g
In 1 g of 212Bi, the amount of 208Ti formed = 7/20 g
∴ Amount of 208Ti present in 0.5 g = 720×0.5=0.175 g

In 20 g of 212Bi, the amount of 212Po formed = 13 g
In 1 g of 212Bi, the amount of 212Po formed = 13/20 g
∴ Amount of 212Po present in 0.5 g = 1320×0.5=0.325 g

Answer:

(a) Activity, A0 = 8 × 108 dis/sec

(i) InA1A0=In11.7948=0.389
(ii) InA2A0=In9.16808=0.12362
(iii) InA3A0=In7.44928=-0.072
(iv) InA4A0=In6.26846=-0.244
(v) InA5A = In5.41158=-0.391
(vi) InA6A0 = In3.08288=-0.954
(vii)InA7A0= In91.88998=-1.443
(viii) In1.16718=In90.72128=-1.93
(ix) In0.72128=In90.72128=-2.406

The required graph is given below.


(b) Half-life of 110Ag = 24.4 s
(c) Half-life of 110Ag, T12 = 24.4 s

Decay constant, λ=0.693T12
  λ=0.69324.4=0.0284
t = 50 sec

 Activity, A=A0e-λt                 =8×108×e-0.0284×50                   =1.93×108

(d)

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

Page No 444:

Question 44:

(a) Activity, A0 = 8 × 108 dis/sec

(i) InA1A0=In11.7948=0.389
(ii) InA2A0=In9.16808=0.12362
(iii) InA3A0=In7.44928=-0.072
(iv) InA4A0=In6.26846=-0.244
(v) InA5A = In5.41158=-0.391
(vi) InA6A0 = In3.08288=-0.954
(vii)InA7A0= In91.88998=-1.443
(viii) In1.16718=In90.72128=-1.93
(ix) In0.72128=In90.72128=-2.406

The required graph is given below.


(b) Half-life of 110Ag = 24.4 s
(c) Half-life of 110Ag, T12 = 24.4 s

Decay constant, λ=0.693T12
  λ=0.69324.4=0.0284
t = 50 sec

 Activity, A=A0e-λt                 =8×108×e-0.0284×50                   =1.93×108

(d)

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

Answer:

Given:
Time taken by the body to excrete half the amount, t1 = 24 hours
Half-life of radioactive isotope, t2 = 6 hours
Initial activity, A0 = 6 μCi
Let after time t, activity of the sample be A.
Half-life period T1/2 is given by
 T1/2=t1t2t1+t2=24×624+6         =24×630=4.8 h
Activity A at time t is given by
 A= A02t/T1/2 3μCi=6μCi2t/4.86μCi2t/4.8=3 t=4.8 h

Page No 444:

Question 45:

Given:
Time taken by the body to excrete half the amount, t1 = 24 hours
Half-life of radioactive isotope, t2 = 6 hours
Initial activity, A0 = 6 μCi
Let after time t, activity of the sample be A.
Half-life period T1/2 is given by
 T1/2=t1t2t1+t2=24×624+6         =24×630=4.8 h
Activity A at time t is given by
 A= A02t/T1/2 3μCi=6μCi2t/4.86μCi2t/4.8=3 t=4.8 h

Answer:

Discharging of a capacitor through a resistance R is given by
Q=qe-t/CR
Here, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance

Energy, E = 12Q2C = q2e-2t/CR2C

Activity, A = A0e-λt
Here, A0 = Initial activity
λ = Disintegration constant

∴ Ratio of the energy to the activity = EA=q2×e-2t/CR2CA0e-λt
Since the terms are independent of time, their coefficients can be equated.
2tCR=λt
λ=2CR
1τ=2CR
R=2τC               

 

Page No 444:

Question 46:

Discharging of a capacitor through a resistance R is given by
Q=qe-t/CR
Here, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance

Energy, E = 12Q2C = q2e-2t/CR2C

Activity, A = A0e-λt
Here, A0 = Initial activity
λ = Disintegration constant

∴ Ratio of the energy to the activity = EA=q2×e-2t/CR2CA0e-λt
Since the terms are independent of time, their coefficients can be equated.
2tCR=λt
λ=2CR
1τ=2CR
R=2τC               

 

Answer:

Given:
Resistance of resistor, R = 100 Ω
Inductance of an inductor, L = 100 mH
Current i at any time t is given by
i=i01-e-RtL
Number of active nuclei N at any time t is given by
N=N0e-λt
Where N0 = Total number of nuclei
λ = Disintegration constant
Now,
iN=i01-e-tR/LN0e-λt
As iN is independent of time, coefficients of t are equal.

Let t12 be the half-life of the isotope.

 -RL=-λRL=0.693t12t12=0.693×10-3=6.93×10-4 s

Page No 444:

Question 47:

Given:
Resistance of resistor, R = 100 Ω
Inductance of an inductor, L = 100 mH
Current i at any time t is given by
i=i01-e-RtL
Number of active nuclei N at any time t is given by
N=N0e-λt
Where N0 = Total number of nuclei
λ = Disintegration constant
Now,
iN=i01-e-tR/LN0e-λt
As iN is independent of time, coefficients of t are equal.

Let t12 be the half-life of the isotope.

 -RL=-λRL=0.693t12t12=0.693×10-3=6.93×10-4 s

Answer:

235 g of uranium contains 6.02 × 1023 atoms.
1 g of uranium = 1235×6.023×1023 atoms
∴ 0.7 g of uranium = 1235×6.023×1023 ×0.007 atoms

1 atom gives 200 MeV. 

∴ Total energy released = 6.023×1023×0.007×200×106×1.6×10-19235 J = 5.74×108 J

Page No 444:

Question 48:

235 g of uranium contains 6.02 × 1023 atoms.
1 g of uranium = 1235×6.023×1023 atoms
∴ 0.7 g of uranium = 1235×6.023×1023 ×0.007 atoms

1 atom gives 200 MeV. 

∴ Total energy released = 6.023×1023×0.007×200×106×1.6×10-19235 J = 5.74×108 J

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×200×106×1.6×10-19 = Power
 N×200×106×1.6×10-19 = 300×106 
N=32×1.6×1019 = 33.2×1019  atoms
6.023 × 1023 atoms = 238 gm of U235
33.2×1019 atoms will present in 238×3×10196×1023×3.2=3.7 mg

Page No 444:

Question 49:

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×200×106×1.6×10-19 = Power
 N×200×106×1.6×10-19 = 300×106 
N=32×1.6×1019 = 33.2×1019  atoms
6.023 × 1023 atoms = 238 gm of U235
33.2×1019 atoms will present in 238×3×10196×1023×3.2=3.7 mg

Answer:

(a) Total population of the town = 1 million =  106
Average electric power needed per person = 300 W
Total power used by the town in one day = 300 × 106 × 60 × 60 × 24 J = 300 × 86400 ×106 J
Energy generated in one fission = 200 × 106 × 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
Electrical energy, E=3.2×10-11×25100E=8×10-12 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 × 106 J
N = 3.24 × 1024

(b) Number of moles required per day n = N6.023×1023
n = 3.24×10246.023×1023 = 5.38 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 235U be used.
 x×3100 = 1.264×30x = 1264 kg

Page No 444:

Question 50:

(a) Total population of the town = 1 million =  106
Average electric power needed per person = 300 W
Total power used by the town in one day = 300 × 106 × 60 × 60 × 24 J = 300 × 86400 ×106 J
Energy generated in one fission = 200 × 106 × 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
Electrical energy, E=3.2×10-11×25100E=8×10-12 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 × 106 J
N = 3.24 × 1024

(b) Number of moles required per day n = N6.023×1023
n = 3.24×10246.023×1023 = 5.38 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 235U be used.
 x×3100 = 1.264×30x = 1264 kg

Answer:

 (a) Q=2×m H12-mH33+mH13c2         =(4.028204-4.023874)×931 MeV         =4.05 MeV(b) Q=2×m H12-(m H23+mn)c2         =[4.028204-4.024694)×931         =0.00351×931         =3.25 MeV(c) Q=m H12+m H13-m He14-mnc2        =(2.014102+3.016049-4.002603-1.008665)×931        =17.57 MeV

Page No 444:

Question 51:

 (a) Q=2×m H12-mH33+mH13c2         =(4.028204-4.023874)×931 MeV         =4.05 MeV(b) Q=2×m H12-(m H23+mn)c2         =[4.028204-4.024694)×931         =0.00351×931         =3.25 MeV(c) Q=m H12+m H13-m He14-mnc2        =(2.014102+3.016049-4.002603-1.008665)×931        =17.57 MeV

Answer:

Given:
Average thermal energy, E = 1.5 kT
Point of coulomb potential energy = 2 fm
Potential energy is given by
U = Kq1q2r   .....(1)
Here, K = 14πε0 = 9×109

Charge, q1=q2=2×1.6×10-19 C  

Average kinetic energy E is given by
 E=32kT     .....(2)
Here, k = Boltzman constant
T = Temperature       

Equating equation (1) and (2), we get
kq1q2r=32kT
T=2Kq1q23kr
     = 2×9×109×4×1.6×10-1923×1.38×10-23×2×10-15
     = 2.23×1010 K

Page No 444:

Question 52:

Given:
Average thermal energy, E = 1.5 kT
Point of coulomb potential energy = 2 fm
Potential energy is given by
U = Kq1q2r   .....(1)
Here, K = 14πε0 = 9×109

Charge, q1=q2=2×1.6×10-19 C  

Average kinetic energy E is given by
 E=32kT     .....(2)
Here, k = Boltzman constant
T = Temperature       

Equating equation (1) and (2), we get
kq1q2r=32kT
T=2Kq1q23kr
     = 2×9×109×4×1.6×10-1923×1.38×10-23×2×10-15
     = 2.23×1010 K

Answer:

Given:
Atomic mass of 8Be = 8.0053 u
Atomic mass of 4He = 4.0026 u

Required Q-value=(2×4.0026-8.0053) c2                                  =0.0001×931 MeV                                  =0.0001×931×106 eV                                  =93.1 KeV

No, such reaction is not favourable.

Page No 444:

Question 53:

Given:
Atomic mass of 8Be = 8.0053 u
Atomic mass of 4He = 4.0026 u

Required Q-value=(2×4.0026-8.0053) c2                                  =0.0001×931 MeV                                  =0.0001×931×106 eV                                  =93.1 KeV

No, such reaction is not favourable.

Answer:

Given:
18 g of water contains 6.023 × 1023 molecules.

∴ 1000 g of water =6.023×1023×100018=3.346×1025 molecules
 % of deuterium = 3.346×1025×0.015100 = 0.05019 × 1023
Energy of deuterium = 30.4486×1025
                               =2×mH2-mH3-mpc2=2×2.014102 u-3.016049 u-1.007276 uc2=0.004879×931 MeV=4.542349 MeV=7.262 ×10-13 J
    Total energy = 0.05019 × 1023× 7.262 × 10-13 J
                               = 3644 MJ
                             



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