Thinking Process

$T_{\sqrt{ } 2}=\frac{\ln 2}{\lambda}, \lambda \rightarrow$ decay constant

Answer

(c) Radioactivity is a process due to which a radioactive material spontaneously decays. In half-life $(t=1 yr)$ of the material on the average half the number of atoms will decay.

Therefore, the containers will in general have different number of atoms of the material, but their average will be approx 5000 .

Answer

(b) Given, $F=\frac{G M m}{r^{2}}$

$$ \begin{aligned} M & =\text { effective mass of hydrogen atom } \\ & =\text { mass of electron }+ \text { mass of proton }-\frac{B^{2}}{C} \end{aligned} $$

where $B$ is $B E$ of hydrogen atom $=13.6 eV$.

Answer

(b) $\alpha-\beta$ particle carries one unit of negative charge, an $\alpha$-particle carries 2 units of positive charge and $\gamma$ (particle) carries no charge, therefore electronic energy levels of the atom charges for $\alpha$ and $\beta$ decay, but not for $\gamma$-decay.

Answer

(a) Let the nucleus is $ _{z} X^{A} \cdot \beta^{+}$decay is represented as

$${}_z X^A \rightarrow _{z-1} Y^A+ _ {+1} e^0 +v+Q_2 \newline $$

$$\therefore \quad Q_2 =[m( {} _z X^A) - m_n ({} _{z-1} Y^A) - m _{e}] c^{2} $$

$$ =[m( {} _z X^A) + zm _e - m_n ({} _{z-1} Y^A) -(z-1) m_e - 2m_e]c^{2} $$

$$ =[m( {} _z X^A) -m({} _{z-1} Y^A)-2 m _{e}] c^{2} $$

$$ =(M _{x}- M _{y} -2 m _{e}) c^{2} $$

$\beta^{-}$decay is represented as

$$ \therefore \quad {}_z X^A \rightarrow _{z-1} y^A+ _ {+1} e^0 +v+Q_2 \newline $$

$$ \alpha_1 =[m( {} _z X^A) - m_n ({} _{z-1} Y^A) - m _{e}] c^{2} $$

$$ =[m( {} _z X^A) + zm _e - m_n ({} _{z+1} Y^A) -(z-1) m_e ] c^2 $$

$$ =[m( {} _z X^A) -m({} _{z-1} Y^A)] c^{2} $$

$$ =(M _{x}- M _{y} ) c^{2} $$

Thinking Process

Isotopes of an element are having same atomic numbers and different mass numbers.

Answer

(a) Tritium $( _1 H^{3})$ contains 1 proton and 2 neutrons. A neutron decays as $n \rightarrow P+\bar{e}+\bar{v}$, the nucleus may have 2 protons and one neutron, i.e., tritium will transform into $2 He^{3}$ (2 protons and 1 neutron).

Triton energy is less than that of $2 He^{3}$ nucleus, i.e., transformation is not allowed energetically.

Answer

(b) Stable heavy nuclei have more neutrons than protons. This is because electrostatic force between protons is repulsive, which may reduce stability.

Thinking Process

When there is an elastic collision between two bodies of same mass their velocities are exchanged.

Answer

(b) According to the question, the moderator used have light nuclei (like proton). When protons undergo perfectly elastic collision with the neutron emitted their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons. Heavy nuclei will not serve the purpose because elastic collisions of neutrons with heavy nuclei will not slow them down.

Answer

$(a, b)$

Fusion processes are impossible at ordinary temperatures and pressures. The reason is nuclei are positively charged and nuclear forces are short range strongest forces.

Answer

$(b, d)$

The two samples of the two radioactive nuclides $A$ and $B$ can simultaneously have the same decay rate at any time if initial rate of decay of $A$ is twice the initial rate of decay of $B$ and $\lambda_{A}>\lambda_{B}$. Also, when initial rate of decay of $B$ is same as rate of decay of $A$ at $t=2$ hand $\lambda_{B}<\lambda_{A}$.

Answer

$(c, d)$

From the given figure, it is clear that slope of curve $A$ is greater than that of curve $B$. So rate of decay is faster for $A$ than that of $B$.

We know that $(\frac{d N}{d t}) \propto, \lambda$, at any instant of time hence we can say that $\lambda_{A}>\lambda_{B}$. At point $P$ shown in the diagram the two curve intersect.

Hence at point $P$, rate of decay for both $A$ and $B$ is the same.

Answer

Nuclei $He_2^{3}$ and $He_1^{3}$ have the same mass number. $He_2^{3}$ has two proton and one neutron. $He_1^{3}$ has one proton and two neutron. The repulsive force between protons is missing in $ _1 He^{3}$ so the binding energy of $ _1 He^{3}$ is greater than that of $ _2 He^{3}$.

Answer

We know that, rate of decay $=\frac{-d N}{d t}=\lambda N$

where decay constant $(\lambda)$ is constant for a given radioactive material. Therefore, graph between $N$ and $\frac{d N}{d t}$ is a straight line as shown in the diagram.

Answer

From the given figure, we can say that

$$ \text { at } t=0, (\frac{d N}{d t})_{A}=(\frac{d N}{d t}) _{B}\newline \Rightarrow \qquad(N_0)_A=(N_0)_B $$

Considering any instant $t$ by drawing a line perpendicular to time axis, we find that $(\frac{d N}{d t})_A>(\frac{d N}{d t})_B$

$$ \begin{matrix} \Rightarrow & \lambda_{A} N_{A}>\lambda_{B} N_{B} & \\ \because & N_{A}>N_{B} & \text { (rate of decay of } B \text { is slower) } \\ \therefore & \lambda_{B}>\lambda_{A} & \\ \Rightarrow & \tau_{A}>\tau_{B} & [\because \text { Average life } \tau=\frac{1}{\lambda}] \end{matrix} $$

Answer

Excited electron cannot emit radiation because energy of electronic energy levels is in the range of $eV$ and not $MeV$ ( mega electron volt ). $\gamma$-radiations have energy of the order of $MeV$.

Answer

In pair annihilation, an electron and a positron destroy each other to produce $2 \gamma$ photons which move in opposite directions to conserve linear momentum. The annihilation is shown below $ _0 e^{-1}+ _0 e^{+1}\rightarrow 2 \gamma$ ray photons.

Answer

Because protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.

Thinking Process

Based or decay law of unstable radioactive nuclei.

Answer

Consider the situation shown in the graph.

At $t=0, N_{A}=N_0$ (maximum) while $N_{B}=0$. As time increases, $N_{A}$ decreases exponentially and the number of atoms of $B$ increases. They becomes $(N_{B})$ maximum and finally drop to zero exponentially by radioactive decay law.

Thinking Process

Carbon dating is a technique that uses the decay of carbon $-14({ }^{14} C)$ to estimate the age of organic materials, such as wood and leather.

Answer

Given, $R=12 dis / min$ per $g, R_0=16 dis / min$ per $g, T_{1 / 2}=5760 yr$ Let $t$ be the span of the tree

According to radioactive decay law,

$$ R=R_0 e^{-\lambda t} \text { or } \frac{R}{R_0}=e^{-\lambda t} \text { or } e^{\lambda t}=\frac{R_0}{R} $$

Taking log on both the sides

$$ \lambda t \log _e e =\log _e \frac{R_0}{R} \Rightarrow \lambda t=\log _{10} \frac{16}{12} \quad \times 2.303 \newline $$

$$ t =\frac{2.303(\log 4-\log 3)}{\lambda} \newline =\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad [{\because \lambda=\frac{0.6931}{T_{1 / 2}}}] \newline =2391.20 \text{yr} $$

Thinking Process

We have to use de-Broglie formula $(\boldsymbol{\lambda}=h / p)$ to find momentum of the particle.

Answer

Each particle (neutron and proton) present inside the nucleus is called a nucleon.

Let $\boldsymbol{\lambda}$ be the wavelength $\boldsymbol{\lambda}=10^{-15} m$

To detect separate parts inside a nucleon, the electron must have wavelength less than $10^{-15} m$.

We know that

Energy

$$ \begin{aligned} \lambda & =\frac{h}{p} \text { and } KE=PE \\ & =\frac{h c}{\lambda} \end{aligned} $$

From Eq. (i) and Eq. (ii),

$$ \begin{aligned} \text { kinetic energy of electron } & =P E=\frac{h c}{\lambda}-\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-15} \times 1.6 \times 10^{-19}} eV \\ KE & =10^{9} eV \end{aligned} $$

Thinking Process

Based on the mirror isobar concept and binding energy concept.

Answer

(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2 , if $Z_1=N_2$ and $Z_2=N_1$.

Now in $ _{11} Na^{23}, Z_1=11, N_1=23-11=12$

$\therefore$ Mirror isobar of $ _{11} Na^{23}$ is $ _{12}, Mg^{23}$, for which $Z_2=12=N_1$ and $N_2=23-12=11=Z_1$

(b) As $ _12^{23} Mg$ contains even number of protons (12) against $ _11^{23} Na$ which has odd number of protons (11), therefore $ _11^{23} Mg$ has greater binding energy than $ _{11} Na^{23}$.

Thinking Process

To solve this problem concept of chain of two decays will be used. For the process,

$A \rightarrow B \rightarrow C$, the law of decay is $\frac{d N_{B}}{d t}=\lambda_{B} N_{B}+\lambda_{A} N_{A}$.

Answer

Consider the chain of two decays

$$ { }^{38} \mathrm{~S} \xrightarrow[2.48 \mathrm{~h}]{ }{ }^{38} \mathrm{Cl} \xrightarrow[0.62 \mathrm{~h}]{ }^{38} \mathrm{Ar} $$

At time $t$, Let ${ }^{38} S$ have $N_1(t)$ active nuclei and ${ }^{38} Cl$ have $N_2(t)$ active nuclei.

Also,

$$ \begin{aligned} \frac{d N_1}{d t} & =-\lambda_1 N_1=\text { rate of formation of } Cl^{38} . \\ \frac{d N_2}{d t} & =-\lambda_1 N_2+\lambda_1 N_1 \\ N_1 & =N_0 e^{-\lambda_1 t} \\ \frac{d N_2}{d t} & =\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2 \end{aligned} $$

$$ \text { But } \quad N_1=N_0 e^{-\lambda_1 t} $$

Multiplying by $e^{\lambda z^{\star}} d t$ and rearranging

$$ e^{\lambda_2 t} d N_2+\lambda_2 N_2 e^{\lambda_2 t} d t=\lambda_1 N_0 e^{(\lambda_2-\lambda_1) t} d t $$

Integrating both sides

$$ \begin{aligned} & N_2 e^{\lambda_2 t}=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} e^{(\lambda_2-\lambda_1) t}+C \\ & \text { Since, } \\ & \text { at } t=0, N_2=0, C=-\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} \\ & N_2 e^{\lambda_2 t}=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}(e^{(\lambda_2-\lambda_1) t}-1) \\ & N_2=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}(e^{-\lambda_1 t}-e^{-\lambda_2 t}) \\ & \frac{d N_2}{d t}=0 \\ & \text { For maximum count, } \quad \text {.. (ii) } \\ & \lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2=0 \\ & e^{\lambda_2 t}-\frac{\lambda_2}{\lambda_1} \cdot \frac{\lambda_1}{N_2}=\frac{\lambda_2}{\lambda_1} e^{\lambda_1 t} \\ & e^{\lambda_2 t}-\frac{.\lambda_2)}{\lambda_2} e^{\lambda_1 t}[e^{(\lambda_2-\lambda_1) t}-1]=0 \text { [From Eq. (i)] } \\ & \text { or } e^{\lambda_2 t}+\frac{\lambda_2}{(\lambda_2-\lambda_1)} e^{\lambda_1 t}=0 \end{aligned} $$

$$ 1-\frac{\lambda_2}{(\lambda_2-\lambda_1)}+\frac{\lambda_2}{(\lambda_2-\lambda_1)} e^{(\lambda_1-\lambda_2) t}=0 \newline $$ $$\frac{\lambda_2}{(\lambda_2-\lambda_1)} e^{(\lambda_1-\lambda_2) t}=\frac{\lambda_2}{(\lambda_2-\lambda_1)}-1 \newline$$

$$e^{(\lambda_1-\lambda_2) t}=\frac{\lambda_1}{\lambda_2} \newline t=\log _{e} \frac{\lambda_1}{\lambda_2} /(\lambda_1-\lambda_2) \newline$$ $$=\frac{\log _e \frac{2.48}{0.62}}{2.48-0.62} \newline $$ $$=\frac{\log _e 4}{1.86}=\frac{2.303 \times 2 \times 0.3010}{1.86} \newline $$

$$=0.745 s \qquad [\because \lambda=\frac{0.693}{T_{1 / 2}}] $$

Note Do not apply directly the formula of radioactive. Apply formulae related to chain decay.

Thinking Process

Apply conservation of energy as well as conservation of momentum.

Answer

Given binding energy

$$ B=2.2 MeV $$

From the energy conservation law,

$$ E-B=K_{n}+K_{p}=\frac{p_n^{2}}{2 m}+\frac{p_p^{2}}{2 m} \qquad …(i) $$

From conservation of momentum,

$$ p_{n}+p_{p}=\frac{E}{C} \qquad …(ii) $$

As

$$ E=B \text {, Eq. (i) } p_n^{2}+p_p^{2}=0 $$

It only happen if $p_{n}=p_{p}=0$

So, the Eq. (ii) cannot satisfied and the process cannot take place.

Let $E=B+X$, where $X«B$ for the process to take place.

Put value of $p_{n}$ from Eq. (ii) in Eq. (i), we get

$$ X=\frac{(\frac{E}{c}-p_{p})}{2 m}+\frac{p_p^{2}}{2 m} $$

$$ or \qquad 2 p_p^{2}-\frac{2 E p_{p}}{c}+\frac{E^{2}}{c^{2}}-2 m X=0 $$

Using the formula of quadratic equation, we get

$$ p_{p}=\frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^{2}}{c^{2}}-8 (\frac{E^{2}}{c^{2}}-2 m X})}{4} $$

For the real value $p_{p}$, the discriminant is positive

$$ \begin{aligned} \frac{4 E^{2}}{c^{2}} & =8 (\frac{E^{2}}{c^{2}}-2 m X) \\ 16 m X & =\frac{4 E^{2}}{c^{2}} \\ X & =\frac{E^{2}}{4 m c^{2}} \approx \frac{B^{2}}{4 m c^{2}} \qquad[\because X < < B \Rightarrow E \cong B] \end{aligned} $$

Answer

The binding energy is $H$-atom

$$ E=\frac{m e^{4}}{\pi \varepsilon_0^{2} h^{2}}=13.6 eV \quad …(i) $$

If proton and neutron had charge $e^{\prime}$ each and were governed by the same electrostatic force, then in the above equation we would need to replace electronic mass $m$ by the reduced mass $m^{\prime}$ of proton-neutron and the electronic charge $e$ by $e^{\prime}$.

$$ \begin{aligned} m^{\prime} & =\frac{M \times N}{M+N}=\frac{M}{2} \\ & =\frac{1836 m}{2}=918 m \end{aligned} $$

Here, $M$ represents mass of a neutron/proton

$$ \therefore \quad \text { Binding energy }=\frac{918 m(e^{\prime})^{4}}{8 \varepsilon_0^{2} h^{2}}=2.2 MeV \quad …(ii) $$

Dividing Eqs. (ii) and (i), we get

$$ \begin{aligned} 918 (\frac{e^{\prime}}{e})^4 & =\frac{2.2 MeV}{13.6 eV}=\frac{2.2 \times 10^{6}}{13.6} \\ (\frac{e^{\prime}}{e})^4 & =\frac{2.2 \times 10^{6}}{13.6 \times 918}=176.21 \\ \frac{e^{\prime}}{e} & =(176.21)^{1 / 4}=3.64 \end{aligned} $$

Answer

Before $\beta$-decay, neutron is at rest. Hence, $E_{n}=m_{n} c^{2}, p_{n}=0$

$$p_{n}=p_{p}+p_{e}$$

$$Or \qquad p_{p} + p_{e} = 0 \Rightarrow |p_p|=|p_{e}|=p $$

Also,

$E_{p} =(m_p^{2} c^{4}+p_p^{2} c^{2})^{\frac{1}{2}}, \\ 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}} $

From conservation of energy,

$$ \begin{aligned} (m_p^{2} c^{4}+p^{2} c^{2})^{\frac{1}{2}}+ & =(m_e^{2} c^{4}+p^{2} c^{2})^{\frac{1}{2}}=m_{n} c^{2} \\ m_{p} c^{2} \approx 936 MeV, m_{n} c^{2} & \approx 938 MeV, m_{e} c^{2}=0.51 MeV \end{aligned} $$

Since, the energy difference between $n$ and $p$ is small, $p c$ will be small, $p c«<m_{p} c^{2}$, while $p c$ may be greater than $m_{e} c^{2}$

$$ \Rightarrow \quad m_{p} c^{2}+\frac{p^{2} c^{2}}{2 m_p^{2} c^{4}}=m_{n} c^{2}-p c $$

To first order $\quad p c \simeq m_{n} c^{2}-m_{p} c^{2}=938 MeV-936 MeV=2 MeV$

This gives the momentum of proton or neutron. Then,

$$ \begin{aligned} E_{p} & =(m_p^{2} c^{4}+p^{2} c^{2})^{\frac{1}{2}}=\sqrt{936^{2}+2^{2}} \\ & \simeq 936 MeV \\ E_{e} & =(m_e^{2} c^{4}+p^{2} c^{2})^{\frac{1}{2}}=\sqrt{(0.51)^{2}+2^{2}} \\ & \simeq 2.06 MeV \end{aligned} $$

Thinking Process

Based on Decay law and half-life.

Answer

In the table given below, we have listed values of $R(MB_{q})$ and $\ln \frac{R}{R_0}$.

(i) When we plot the graph of $R$ versust, we obtain an exponential curve as shown.

From the graph we can say that activity $R$ reduces to $50 %$ in $t=O B \approx 40 min$ So, $t_{1 / 2} \approx 40 min$.

(ii) The adjacent figure shows the graph of $\ln (R / R_0)$ versus $t$.

Slope of this graph $=-\lambda$

from the graph,

Half-life

$$ \begin{aligned} \lambda & =-\frac{-4.16-3.11}{1} \Rightarrow=1.05 h^{-1} \\ T_{1 / 2} & =\frac{0.693}{\lambda}=\frac{0.693}{1.05}=0.66 h \\ & =39.6 min \approx 40 min \end{aligned} $$

Answer

(i) The proton separation energy is given by

$$ \begin{aligned} S_{p S n} & =(M_{119.70}+M_{H}-M_{120,70}) c^{2} \\ & =(118.9058+1.0078252-119.902199) c^{2} \\ & =0.0114362 c^{2} \\ S_{p S p} & =(M_{120,70}+M_{H}-M_{121,70}) c^{2} \\ & =(119.902199+1.0078252-120.903822) c^{2} \\ & =0.0059912 c^{2} \end{aligned} $$

Similarly

Since, $S_{pSn}>S_{pSb}$, Sn nucleus is more stable than Sb nucleus.

(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.