Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 13 - Nuclei

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Page No 81:

Question 1:

Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year,
(a) all the containers will have 5000 atoms of the material.
(b) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000.
(c) the containers will in general have different numbers of the atoms of the material but their average will be close to 5000.
(d) none of the containers can have more than 5000 atoms.

Answer:

Radioactivity is a process in which radioactive material spontaneously decays. The half life of a radioactive material refers to the time taken by radioactive substance to reach half of the original value of the material.
After one half life, the container will in general have different number of atoms but average will be approximately 5000. This is because when half life is completed atoms may or may not be exactly equal to half of the initial number but their average will be closer to half of the initial number.

Hence, the correct answer is option (c).

Page No 81:

Question 2:

The gravitational force between a H-atom and another particle of mass m will be given by Newton’s law:

$F = G \frac{M . m}{r^{2}}, where r is in km and$
(a) M = m_proton+ m_electron^.
(b) M = m_proton + m_{electron $- \frac{B}{c^{2}}$}(B = 13.6 eV).
(c) M is not related to the mass of the hydrogen atom.
(d) M = m_proton + m_electron $- \frac{|V|}{c^{2}}$ (|V| = magnitude of the potential energy of electron in the H-atom).

Answer:

Gravitational force, $F = \frac{G M m}{r^{2}}$
in the given expression M is the effective mass of the hydrogen atom.
During formation of H-atom some mass of nucleons convert into energy by E = mc², this energy is used to bind the nucleons along with nucleus.
So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons
In the given expression M is the effective mass of the hydrogen atom which is given as:
M_effective = m_proton + m_electron – âˆ†m.
Here, âˆ†m is the mass which is related to the binding energy of hydrogen atom. This energy is required for binding proton and electron and forms H-atom.
$B = ∆ m c^{2} [∵ E = m c^{2}] \Rightarrow ∆ m = \frac{B}{c^{2}} (B = 13.6 ev for H - atom) \Rightarrow M_{effective} = m_{proton} + m_{electron} - \frac{B}{c^{2}}$
Hence, the correct answer is option (b).

Page No 82:

Question 3:

When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom

(a) do not change for any type of radioactivity .
(b) change for α and β radioactivity but not for γ-radioactivity.
(c) change for α-radioactivity but not for others.
(d) change for β-radioactivity but not for others.

Answer:

.In α-decay, the mass number of the product nucleus is four less than that of the decaying nucleus (parent nucleus), while the atomic number Z decreases by two. While in both β⁻ and β⁺ decay, the mass number remains unchanged. In β⁻decay, the atomic number Z of the nucleus goes up by 1, while in β⁺ decay Z goes down by 1.

During both alpha and beta decay, ejection of charged particles from the nucleus, altering its charge and its interactions with electrons, which in turn alter the electronic energy levels of the atom.
Whereas in gamma decay, there is no change of charge on the atom, and hence the electronic levels of the atoms are not affected.
Hence, the correct answer is option (b).

Page No 82:

Question 4:

M_x and M_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β^–decay is Q₁ and that for a β ⁺ decay is Q₂. If me denotes the mass of an electron, then which of the following statements is correct?
(a) Q₁ = (M_x – M_y) c² and Q₂ = (M_x – M_y– 2m_e)c²
(b) Q₁ = (M_x– M_y) c² and Q₂ = (M_x– M_y)c²
(c) Q₁ = (M_x – M_y – 2m_e) c² and Q₂ = (M_x – M_y +2 m_e)c²
(d) Q₁ = (M_x – M_y + 2m_e) c²and Q² = (M_x – M_y +2 m_e)c²

Answer:

Let the nucleus be ${}_{Z}^{A}X$ and undergoes β^– decay is represented by:

$\Rightarrow {}_{Z}^{A}X \to {}_{Z + 1}^{A}Y + {}_{- 1}^{0}e + ν Q value = ∆ m c^{2} Here ∆ m is mass defect$
$\begin{array}{rcl} ∆ m & = & [M_{X} - Z m_{e}] - [M_{Y} - (Z + 1) m_{e} + m_{e}] \\ = & M_{X} - Z m_{e} - M_{Y} + Z m_{e} + m_{e} - m_{e} \end{array}$

$\Rightarrow ∆ m = M_{X} - M_{Y} \Rightarrow Q_{1} = (M_{X} - M_{Y}) c^{2} For β_{}^{+} decay$
${}_{Z}^{A}X \to {}_{z - 1}^{A}Y + {}_{+ 1}^{0}e + ν' \Rightarrow ∆ m = [M_{X} - Z m_{e}] - [M_{Y} - (Z - 1) m_{e} + m_{e}] = M_{X} - M_{Y} - 2 m_{e} \Rightarrow Q_{2} = (M_{X} - M_{Y} - 2 m_{e}) c^{2}$

Hence, the correct answer is option (a).

Page No 82:

Question 5:

Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into p + $\bar{e}$ + $\bar{v}$ . If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because

(a) Triton energy is less than that of a He³ nucleus.
(b) the electron created in the beta decay process cannot remain in the nucleus.
(c) both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He³ nucleus.
(d) because free neutrons decay due to external perturbations which is absent in a triton nucleus.

Answer:

Tritium ( ${}_{1}^{3}H$ ) is radioactive isotope of hydrogen which contains 1 proton and 2 neutrons.

And, ( ${}_{2}^{3}{He}$ ) contains 2 protons and 1 neutrons. Repulsion between two protons in ${}_{2}^{3}{He}$ makes its energy higher than ${}_{1}^{3}H .$

Binding energy of ${}_{1}^{3}H$ is much smaller than ${}_{2}^{3}{He}$ , so transformation is not possible energetically.

Hence, the correct answer is option (a).

Page No 82:

Question 6:

Heavy stable nuclei have more neutrons than protons. This is because of the fact that

(a) neutrons are heavier than protons.
(b) electrostatic force between protons are repulsive.
(c) neutrons decay into protons through beta decay.
(d) nuclear forces between neutrons are weaker than that between protons.

Answer:

Electrostatic force between proton-proton is repulsive which causes the unstability of nucleus. So in stable nucleus, neutrons are larger than protons.
Hence, the correct answer is option (b).

Page No 82:

Question 7:

In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because

(a) they will break up.
(b) elastic collision of neutrons with heavy nuclei will not slow them down.
(c) the net weight of the reactor would be unbearably high.
(d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature.

Answer:

Collision occurs between the fast moving neutrons and the moderators in the reactor. The purpose of the collision is to slow down the fast moving neutrons. The heavy moderators cannot serve this purpose because they will not be able to slow down the neutrons during the time of elastic collision. That is why, the moderator used must have light nuclei.

Hence, the correct answer is option (b).

Page No 83:

Question 8:

Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact:

(a) nuclear forces have short range.
(b) nuclei are positively charged.
(c) the original nuclei must be completely ionized before fusion can take place.
(d) the original nuclei must first break up before combining with each other.

Answer:

In fusion processes, like combining two deuterons to form a He nucleus, their positively charged nuclei come closer to nuclear range. So, the electrostatic force increases and to overcome this we need very high temperature and pressure. Also, nuclear forces are short range forces.
Hence the correct options are (a) and (b).

Page No 83:

Question 9:

Samples of two radioactive nuclides A and B are taken. λ_A and λ_B are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?

(a) Initial rate of decay of A is twice the initial rate of decay of B and λ_A = λ_B.
(b) Initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B.
(c) Initial rate of decay of B is twice the initial rate of decay of A and λ_A > λ_B.
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λ_B < λ_A.

Answer:

Rate of decay ,
$|\frac{d N}{d t}| = λ N$

(a) ${(\frac{d N}{d t})}_{A} = 2 {(\frac{d N}{d t})}_{B} at t = 0$
$\begin{array}{l} and λ_{A} = λ_{B} \\ λ_{A} {N_{o}}_{A} = 2 λ_{B} {N_{o}}_{B} \\ \Rightarrow {N_{o}}_{A} = 2 {N_{o}}_{B} \\ For {(\frac{d N}{d t})}_{A} = {(\frac{d N}{d t})}_{B} \\ \Rightarrow λ_{A} {N_{o}}_{A} e^{- λ_{A} t} = λ_{B} {N_{o}}_{B} e^{- λ_{B} t} \\ \Rightarrow \frac{λ_{A} {N_{o}}_{A}}{λ_{B} {N_{o}}_{B}} = e^{t (λ_{A} - λ_{B})} \\ \Rightarrow \frac{2 λ_{B} {N_{o}}_{B}}{λ_{B} {N_{o}}_{B}} = e^{t (λ_{A} - λ_{B})} \\ \Rightarrow 2 = e^{t (λ_{A} - λ_{B})} \\ Since λ_{A} = λ_{B} \Rightarrow λ_{A} - λ_{B} = 0 \Rightarrow e^{t (λ_{A} - λ_{B})} = 1 \\ And, 2 \neq 1 Hence, option (a) is incorrect. \end{array}$

$\begin{array}{l} (b) . {(\frac{d N}{d t})}_{A} = 2 {(\frac{d N}{d t})}_{B} and λ_{A} > λ_{B} \\ λ_{A} {N_{o}}_{A} = 2 λ_{B} {N_{o}}_{B} \\ For λ_{A} {N_{o}}_{A} e^{- λ_{A} t} = λ_{B} {N_{o}}_{B} e^{- λ_{B} t} \\ \Rightarrow 2 = e (λ_{A} - λ_{B}) t \\ \Rightarrow (λ_{A} - λ_{B}) t = \ln 2 (Correct) \end{array}$
$\begin{array}{l} (c) . 2 λ_{A} N o_{A} = λ_{B} N o_{B} and λ_{A} > λ_{B} \\ \Rightarrow \frac{1}{2} = e^{(λ_{A} - λ_{B}) t} \\ \Rightarrow (λ_{A} - λ_{B}) t = - \ln 2 (Incorrect) \end{array}$

$\begin{array}{l} (d) . λ_{B} {N_{o}}_{B} = λ_{A} {N_{o}}_{A} e^{- λ_{A} \times 2} and λ_{B} < λ_{A} \\ \Rightarrow λ_{B} {N_{o}}_{B} = λ_{A} {N_{o}}_{A} e^{- λ_{A} \times 2} \\ \Rightarrow For {(\frac{d N}{d t})}_{A} = {(\frac{d N}{d t})}_{B} \\ \Rightarrow λ_{A} N_{o A} e^{- λ_{A} t} = λ_{A} {N_{o}}_{A} e^{- λ_{A} \times 2} e^{- λ_{B} t} \\ \Rightarrow e^{- 2 λ_{A}} = e^{(λ_{A} - λ_{B}) t} \\ (Inorrect) \end{array}$

Hence the correct answer is option (b) .

Page No 83:

Question 10:

The variation of decay rate of two radioactive samples A and B with time is shown in Fig. 13.1.

Which of the following statements are true?

(a) Decay constant of A is greater than that of B, hence A always decays faster than B.
(b) Decay constant of B is greater than that of A but its decay rate is always smaller than that of A.
(c) Decay constant of A is greater than that of B but it does not always decay faster than B.
(d) Decay constant of B is smaller than that of A but still its decay rate becomes equal to that of A at a later instant.

Answer:

Rate of decay ,
$|\frac{d N}{d t}| = λ N$ , where, $λ$ is decay constant.
So, decay constant of A is greater than that of B but it does not always decay faster than B as shown in graph.
Both the carves A and B intersect each other, so they have same rate of decay at that point.
Hence, the correct options are (c) and (d).

Page No 84:

Question 11:

$H e_{2}^{3}$ and $H e_{1}^{3}$ nuclei have the same mass number. Do they have the same binding energy?

Answer:

The binding energy of $H e_{1}^{3}$ is greater than that of $H e_{2}^{3}$ .
$H e_{2}^{3}$ has 2 protons and 1 neutron, whereas $H e_{1}^{3}$ has 1 proton and 2 neutron. Repulsive force between protons in $H e_{1}^{3}$ is absent, hence the binding energy of $H e_{1}^{3}$ is greater than that of $H e_{2}^{3}$ .

Page No 84:

Question 12:

Draw a graph showing the variation of decay rate with number of active nuclei.

Answer:

Rate of decay = $\frac{d N}{d t} = - λ N$
Here $λ$ is decay constant and N is number of active nuclei.

Page No 84:

Question 13:

Which sample, A or B shown in given figure has shorter mean-life?

Answer:

From the graph, at t = 0
${(\frac{d N}{d t})}_{A} = {(\frac{d N}{d t})}_{B} (∵ - \frac{d N}{d t} = λ N)$
$\Rightarrow λ_{A} {(N_{o})}_{A} = λ_{B} {(N_{o})}_{B} . . . (1)$
Let at any instant t >0,
${(\frac{d N}{d t})}_{A} > {(\frac{d N}{d t})}_{B}$
$\Rightarrow λ_{B} > λ_{A} (∵ N = N_{o} e^{- λ t})$
Mean life, $τ = \frac{1}{λ}$
â€‹ $\Rightarrow τ_{B} < τ_{A}$
Hence, mean life of sample B is shorter than mean life of sample A.

Page No 84:

Question 14:

Which one of the following cannot emit radiation and why?
Excited nucleus, excited electron.

Answer:

Excited nucleus can emit radiation, but excited electron cannot. Excited electrons can not emit radiations, because energy of electrons in the energy levels is of order of eV, whereas gamma radiations emitted by excited nucleus have energy is of the order of 10⁶ times eV.

Page No 84:

Question 15:

In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved?

Answer:

When an electron and position destroy each other they produce $γ$ photons which will move in opposite direction and they conserve momentum .
The annihilation is given by.
$o^{e^{- 1}} + o^{e^{+ 1}} \to 2 γ$

Page No 84:

Question 16:

Why do stable nuclei never have more protons than neutrons?

Answer:

A proton has positive charge of 16 × 10^–19 C where as neutron has no charge which means neutrons are neutral. Also, protons repel each other due to electrostatic force of repulsion as charges of same nature repel each other.
In nucleus, nucleons means protons are neutrons are present. The nuclear forces are short range, strong forces which are independent of charge. The strong force between n – n, n – p, p – p are hold inside the nucleus. If number of protons will be more them they will repel each other due to electrostatic force of repulsion and it will create a situation for unstable nucleus. In order to attain a stability, number of protons should be less than neutrons.

Page No 84:

Question 17:

Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence:
A → B → C

Here B is an intermediate nuclei which is also radioactive. Considering that there are N₀ atoms of A initially, plot the graph showing the variation of number of atoms of A and B versus time.

Answer:

Given:- A → B → C, where A is radioactive and C is stable nucleus.
Let us consider A have no atoms initially or at t = 0, N₀. As time increases N_A will decrease in exponential way whereas number of atoms of B will also increases. Being B an intermediate nucles, it is also radioactions so it will start decaying and then drops to zero eaponentially as per law of radioactive decay.

Page No 84:

Question 18:

A piece of wood from the ruins of an ancient building was found to have a ¹⁴C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ¹⁴C is 5760 years.

Answer:

Given → Rate of disintegration, R = 12 dis/min/g initial disintegration rate, R_o= 16 atoms per ninper gram.
Half life, $T_{\frac{1}{2}} = 5760 year .$
Applying radioactive declay law
$N = N_{o} e^{- λ t} or, R = R_{o} e^{^{- λ t}} \Rightarrow 12 = 16 e^{^{- λ t}} \Rightarrow e^{^{- λ t}} = \frac{12}{16} or, e^{^{λ t}} = \frac{16}{12} \Rightarrow \log_{e} e^{^{λ t}} = \log_{e} \frac{4}{3} \Rightarrow λ t = \log_{e} \frac{4}{3} \Rightarrow t = \frac{2.303 \times \log_{10} (\frac{4}{3})}{λ} = \frac{2.303 (\log 4 - \log 3)}{λ} [λ = \frac{0.6931}{T_{\frac{1}{2}}}] \Rightarrow t = \frac{2.303 (0.6020 - 4.771) \times 5760}{0.6931} \Rightarrow t = 2391.2 year .$
Hence, the correct value of t = 2391.2 years

Page No 84:

Question 19:

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10^–15 m.

Answer:

Nucleons consist of neutrons and protons which are present inside the nucleus and are fundamental particles.
In order to resolve two objects which are separated by distance d, then the wavelength of probing signal should be equal to or less than d
Here d ≈ 10^–15 m.
For detecting separate parts, electrons must have wavelength less than 10^–15m.
As wavelength, $λ = \frac{h}{P}$
Kinetic energy, K.E. = PC = $\frac{b c}{λ}$
$\Rightarrow K . E . = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 15}} = 19.89 \times 10^{- 11} J \Rightarrow K . E . = \frac{19.89 \times 10^{- 11}}{1.6 \times 10^{- 19}} \approx 10^{9} ev .$
Hence, the value of kinetic energy is 10⁹eV.

Page No 84:

Question 20:

A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z₁ =N₂and Z₂ =N₁ . (a) What nuclide is a mirror isobar of ${}_{11}^{23}{Na}$ ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?

Answer:

(a). It is given that nuclides 1 is mirror isobar of nuclide 2.
For sodium, ${}_{11}^{23}{Na}$ , Z₁ = 11, N₁ = 23 – 11 Mirror isobar of ${}_{11}^{23}{Na}$ will be ${}_{12}^{23}{Mg},$ here Z₂ = 12 = N₁ and N₂ = 11 = Z₁

(b). The binding energy of ${}_{11}^{23}{Na}$ is 186.6 Me V whereas binding energy of ${}_{12}^{23}{Mg}$ is 181.7 Me V.
B.F_mg < B.E._Na
So, ${}_{11}^{23}{Na}$ has less number of protons than ${}_{12}^{23}{Mg} .$

Page No 85:

Question 21:

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is :

${}^{38}{Sulphur} \to_{= 2.48 h}^{half - life} {}^{38}{Cl} \to_{= 0.62 h}^{half - life} {}^{38}{Ar} (stable)$
Assume that we start with 1000³⁸S nuclei at time t = 0. The number of ³⁸Cl is of count zero at t = 0 and will again be zero at t = ∞ . At what value of t, would the number of counts be a maximum?

Answer:

Given :
${}^{38}S \to_{= 2.48 h}^{half - life} {}^{38}{Cl} \to_{= 0.62 h}^{half - life} {}^{38}{Ar} (stable)$
So, at a given time t, active nuclei in ³⁸S be N₁ (t) and ³⁸Cl be N₂ (t),
$\Rightarrow \frac{{dN}_{1}}{dt} = λ_{1} N_{1} and \frac{{dN}_{2}}{dt} = λ_{2} N_{1} + λ_{1} N_{1} But N_{1} = N_{o} e^{- λ, t} \Rightarrow = \frac{{dN}_{2}}{dt} λ_{1} N_{o} e^{- λ, t} - λ_{2} N_{2}$
Multiply the above equation by $e^{γ_{2} t},$ and on rearranging, $e^{γ_{2} t},$ ${dN}_{2} + λ_{2} N_{2} e^{λ_{2} t} dt = λ_{1} N_{o} e^{(λ_{2} - λ_{1}) t} d$ Intergrading both sides:-
$N_{2} e^{λ_{2} t} = \frac{N_{o} λ_{1}}{λ_{2} - λ_{1}} e^{(λ_{2} - λ_{1}) t} + C At t = 0, N_{2} = 0 \Rightarrow C = \frac{- N_{0} λ_{1}}{λ_{2} - λ_{1}} \Rightarrow N_{2} e^{λ_{2} t} = \frac{- N_{0} λ_{1}}{λ_{2} - λ_{1}} (e^{(λ_{2} - λ_{1}) t} - 1) \Rightarrow N_{2} = \frac{- N_{0} λ_{1}}{λ_{2} - λ_{1}} (e^{- λ_{1} t} - e^{^{- λ_{2} t}}) For maximum, \frac{{dN}_{2}}{dt} = 0 \Rightarrow λ_{1} N_{o} e^{- λ, t} - λ_{2} N_{2} = 0 \Rightarrow \frac{N_{0}}{N_{2}} = \frac{λ_{2}}{λ} e^{λ_{1} t}$
$e^{λ_{2} t} - \frac{λ_{2}}{λ_{2} - λ_{1}} e^{λ_{2} t} + \frac{λ_{2}}{λ_{2} - λ_{1}} e^{λ_{1} t} = 0 \Rightarrow 1 - \frac{λ_{2}}{λ_{2} - λ_{1}} + \frac{λ_{2}}{λ_{2} - λ_{1}} e^{(λ_{1} - λ_{2}) t} = 0 \Rightarrow \frac{λ_{2}}{λ_{2} - λ_{1}} e^{(λ_{1} - λ_{2}) t} = \frac{λ_{2}}{λ_{2} - λ_{1}} - 1 \Rightarrow e^{(λ_{1} - λ_{2}) t} = \frac{λ_{1}}{λ_{2}} \Rightarrow t = \log_{e} (\frac{λ_{1}}{λ_{2}}) (λ_{1} - λ_{2}) \Rightarrow t = \frac{\log_{e} (\frac{2.48}{0.62})}{2.48 - 0.62} = \frac{2.303 \times 2 \times 0.3010}{1.86} \Rightarrow t = 0.74 s$
Hence, the value of t is 0.74 s.

Page No 85:

Question 22:

Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ -ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence calculate how much bigger than B must E be for such a process to happen.

Answer:

Given → Binding energy of deuteron = 2.2 MeV. some part of energy of γ -ray is used up against binding energy of B = 2.2 MeV and the rest part will import K.E. to neutron and proton.
$\Rightarrow E - B = K_{n} + K_{p} \Rightarrow E - B = \frac{P n^{2}}{2 n} + \frac{P p^{2}}{2 n}$
Applying conservation of momentum,
P_n+ P_p = momentum of γ -ray of energy E
$\Rightarrow P_{n} + P_{p} = \frac{E}{c} . . . . . (1)$
When E = B,
then, $\frac{P n^{2} + P p^{2}}{2 m} = 0$
So, Energy E ≠ 0
Now when,
E > B
or E = B + $λ$
$\Rightarrow (B + λ) - B = \frac{P n^{2}}{2 m} + \frac{P n^{2}}{2 m} \Rightarrow λ = \frac{1}{2 m} [{(\frac{E}{c} - P_{p})}^{2} + {P_{p}}^{2}] \Rightarrow P_{p} = \frac{\pm \frac{2 E}{c} \pm \sqrt{\frac{4 E^{2}}{c^{2}} - 4 \times 2 (\frac{E^{2}}{c^{2}} - 2 m λ)}}{4}$
For real and positive value of P_p, discriminant must be zero:
$∴ \frac{4 E^{2}}{C^{2}} - 8 [\frac{E^{2}}{C^{2}} - 2 m λ] = 0 \Rightarrow - E^{2} + 4 m c^{2} λ = 0 \Rightarrow λ = \frac{E^{2}}{4 m c^{2}}$
So, E = B
$\Rightarrow λ \approx \frac{B^{2}}{4 m c^{2}} = \frac{E^{2}}{4 m c^{2}}$

Page No 85:

Question 23:

The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′:

$F = \frac{1}{4 π ε_{0}} \frac{e^{' 2}}{r}$
estimate the value of (e'/e) given that the binding energy of a deuteron is 2.2 MeV.

Answer:

Binding energy of deuteron, B.E. = 2.2 MeV
Binding energy of the atom in ground state,
$E = \frac{m e^{4}}{8 π ε_{0} n^{2}} = 13.6 eV . . . . . (1)$
If the proton and neutron had charge e' each and have same electrostatic force then we can replace m by reduced mass m and e by e.
$m^{1} = \frac{M}{2} = \frac{1836 m}{2} = 918 m . \Rightarrow Binding energy = \frac{918 m {(e^{'})}^{4}}{8 {ε_{0}}^{2} h^{2}} = 2.2 MeV . . . . . (2)$
On dividing equation (2) by (1), we get
$918 {(\frac{e^{'}}{e})}^{4} = \frac{2.2 MeV}{13.6 eV} = \frac{2.2 \times 10^{6}}{13.6} \Rightarrow {(\frac{e^{'}}{e})}^{4} = \frac{2.2 \times 10^{6}}{13.6 \times 918} = 176.21 \Rightarrow \frac{e^{'}}{e} = 3.64$
Hence, the estimated value is 3.64.

Page No 85:

Question 24:

Before the neutrino hypothesis, the beta decay process was throught to be the transition,

$n \to p + \bar{e}$
If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them.Experimentally, the electron energy was found to have a large range.

Answer:

$\Rightarrow m_{p} c^{2} + \frac{p^{2} c^{2}}{2 m p^{2} c^{4}} = m_{n} c^{2} - p c$ Let us consider the cases before and after β-decay.
Before β-decay:- Neutron was at rest.
$\Rightarrow E_{n} = m_{n} c^{2}, P_{n} = 0 After β - decay, P_{n} = P_{p} + P_{e} \Rightarrow 0 = P_{p} + P e \Rightarrow |P_{p}| = |P_{p}| . . . . . (1)$
$Energy of photon, E_{p} = {({m_{p}}^{2} c^{4} + {P_{p}}^{2} c^{2})}^{\frac{1}{2}} Energy of electron, E_{e} = {({m_{e}}^{2} c^{4} + {P_{p}}^{2} c^{2})}^{\frac{1}{2}} = {({m_{e}}^{2} c^{4} + {P_{e}}^{2} c^{2})}^{\frac{1}{2}}$
Applying conservation of energy, ${({m_{p}}^{2} c^{4} + P^{2} c^{2})}^{\frac{1}{2}} = {({m_{e}}^{2} c^{4} + P^{2} c^{2})}^{\frac{1}{2}} = m_{n} c^{2}$
Here, m_pe² = 936 MeV
m_nc² = 938 MeV
m_ec² = 0.15 MeV
Energy difference between p and n is very less, Pc << m_pc²
However pc can be more than m_ec².

$\Rightarrow m_{p} c^{2} + \frac{p^{2} c^{2}}{2 {m_{p}}^{2} c^{4}} = m_{n} c^{2} - p c$
For first order, pc = m_nc² – m_pc²
= 2 MeV.
$\Rightarrow E_{p} {({m_{p}}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} = \sqrt{{(936)}^{2} + {(2)}^{2}} = 936 MeV . Similarly, E_{e} = {({m_{e}}^{2} c^{4} + p^{2} c^{2})}^{\frac{1}{2}} = \sqrt{{(0.51)}^{2} + 22} \approx 2.06 MeV .$

Page No 85:

Question 25:

The activity R of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

t (h)	0	1	2	3	4
R (MBq)	100	35.36	12.51	4.42	1.56

(i) Plot the graph of R versus t and calculate half-life from the graph.
(ii) Plot the graph of ln

(\frac{R}{R_{0}})

versus t and obtain the value of half-life from the graph.

Answer:

(i)

t (h)	0	1	2	3	4
R (MBq)	100	35.36	12.51	4.42	1.56

From the graph when,

R = \frac{R_{0}}{2}, t = T_{\frac{1}{2}}

This means when action when R = 50 MBq
⇒ t = (0.65) (1 hr) = 39 min ≈ 40 min.
(ii) In order to plot the graph of ln

(\frac{R}{R_{0}})

vs t, first we shall calculate ln

(\frac{R}{R_{0}})

for the respective t values.
When t = 1 hr.

\ln (\frac{R}{R_{0}}) = 2.3026 \log (\frac{35.36}{100}) = 2.3026 [\log 35.36 - \log 100] = - 1.0396 \approx - 1.04 .

When t = 2 h

\Rightarrow \ln (\frac{R}{R_{0}}) = 2.3026 \log (\frac{12.51}{100}) = - 2.0788 \approx - 2.08

When t = 3 h
$\Rightarrow \ln (\frac{R}{R_{0}}) = 2.3026 \log (\frac{4.42}{100}) = - 3.1191 \approx 3.12$
When t = 4 h
$\Rightarrow \ln (\frac{R}{R_{0}}) = 2.3026 \log (\frac{1.56}{100}) = \approx - 4.16$
The graph between

\ln (\frac{R}{R_{0}})

versus t will be:

Page No 86:

Question 26:

Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable. (i) Verify this by calculating the proton separation energy S_p for ¹²⁰S_n (Z = 50) and ¹²¹Sb = (Z = 51). The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by

S_p = (M_Z–1, _N + M_H – M_Z,N) c².

Given ¹¹⁹In = 118.9058u, ¹²⁰Sn = 119.902199u, ¹²¹Sb = 120.903824u, ¹H = 1.0078252u.
(ii) What does the existance of magic number indicate?

Answer:

Given expression for proton separation energy,
S_p = (M_Z–1, _N + M_H – M_Z,N) c²
The proton separation energy S_p for ¹²⁰S_n (Z = 50)
S_p _Sn = (M_{49, 70} + M_H – M_{50, 70})c²    (as, N = 120 – 50=70)
          = (118.9085 + 1.0078252 – 119.902199)c²
= 0.0114262 c²
Similarly,
S_p_Sb = (M_{50, 70} + M_H – M_{51, 70})c²                          (as, N = 121 – 51=70)
= (119.902199 + 1.0078252 – 120.903824)c²
= 0.0062002 c²
Sn nucleus will be more stable than Sb nucleus, since  S_p _Sn > S_p_Sb.
(ii) Existence of magic number indicates that the shell structure of nucleus is same as that of shell structure of an atom.
It also explains the peaks in binding energy per nucleon curve.

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