Radioactivity MCQ Questions & Answers in Modern Physics | Physics

Let initially there are total $${N_0}$$ number of nuclei
$$\eqalign{ & {\text{At time }}t\frac{{{N_B}}}{{{N_A}}} = 0.3\left( {{\text{given}}} \right) \cr & \Rightarrow {N_B} = 0.3{N_A} \cr & {N_0} = {N_A} + {N_B} = {N_A} + 0.3{N_A} \cr & \therefore {N_A} = \frac{{{N_0}}}{{1.3}} \cr} $$
As we know $${N_t} = {N_0}{e^{ - \lambda t}}$$
$$\eqalign{ & {\text{or, }}\frac{{{N_0}}}{{1.3}} = {N_0}{e^{ - \lambda t}} \cr & \frac{1}{{1.3}} = {e^{ - \lambda t}} \Rightarrow \ln \left( {1.3} \right) = \lambda t \cr & {\text{or, }}t = \frac{{\ln \left( {1.3} \right)}}{\lambda } \Rightarrow t = \frac{{\ln \left( {1.3} \right)}}{{\frac{{\ln \left( 2 \right)}}{T}}} = \frac{{\ln \left( {1.3} \right)}}{{\ln \left( 2 \right)}}T \cr} $$

Half life, $${t_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda }$$
Mean life, $$\tau = \frac{1}{\lambda }$$
$$\therefore \frac{{{t_{\frac{1}{2}}}}}{\tau } = \ln 2.$$

\[_{72}{A^{180}}{\xrightarrow{\alpha }_{70}}{A_1}^{176}{\xrightarrow{\beta }_{71}}{A_2}^{176}{\xrightarrow{\alpha }_{69}}{A_3}^{172}{\xrightarrow{\gamma }_{69}}{A_4}^{172}\]

Suppose $$x$$ $$\alpha $$-particles and $$y$$ $$\beta $$-particles are emitted
So, change in mass no. is given by
$$\eqalign{ & 4x = 200 - 168 = 32 \cr & x = 8 \cr} $$
and change in atomic no. is given by
$$2x - y = 90 - 80 = 10$$
putting value of $$x$$
$${\text{or}}\,\,2 \times 8 - y = 10$$
So, no. of $$\beta $$-particles $$y = 6$$
No. of $$\alpha $$-particles $$x = 8$$
Alternative
$$\eqalign{ & ^{200}{X_{90}}{ \to ^{168}}{Y_{80}} \cr & {\text{As,}}\,{\,^{200}}{X_{90}} \to \left( {{n_2}H{e^4}} \right) + m\left( {_{ - 1}{\beta ^0}} \right){ + ^{168}}{Y_{80}} \cr} $$
Therefore, in this reaction
$$\eqalign{ & 200 = 4n + 168\,\,{\text{or}}\,\,n = \frac{{200 - 168}}{4} = 8 \cr & {\text{Also,}}\,\,90 = 2n - m + 80 \cr & {\text{or}}\,\,m = 2n + 80 - 90 \cr & = 2 \times 8 + 80 - 90 = 6 \cr} $$
Thus, respective number of $$\alpha $$ and $$\beta $$-particles will be 8 and 6.

Since in the given reaction $$_{ - 1}{e^0}$$  and antineutrino $$\left( {\bar \nu } \right)$$  are released, so it can be considered $$\beta $$-decay.

For $${A_{t\frac{1}{2}}} = 20\,\min, t = 80\,\min,$$      number of half lifes $$n = 4$$
$$\therefore $$ Nuclei remaining = $$\frac{{{N_0}}}{{{2^4}}}.$$  Therefore nuclei decayed $$ = {N_0} - \frac{{{N_0}}}{{{2^4}}}$$
For $${B_{t\frac{1}{2}}} = 40\,\min., t = 80\,\min,$$      number of half lifes $$n = 2$$
$$\therefore $$ Nuclei remaining $$ = \frac{{{N_o}}}{{{2^2}}}.$$  Therefore nuclei decayed $$ = {N_0} - \frac{{{N_0}}}{{{2^2}}}$$
$$\therefore {\text{Required ratio}} = \frac{{{\text{ }}No{\text{ }} - \frac{{{\text{ }}No{\text{ }}}}{{{2^4}}}}}{{{\text{ }}No{\text{ }} - \frac{{{\text{ }}No{\text{ }}}}{{{2^2}}}}} = \frac{{1 - \frac{1}{{16}}}}{{1 - \frac{1}{4}}} = \frac{{15}}{{16}} \times \frac{4}{3} = \frac{5}{4}$$

Average binding energy per nucleon of a nucleus is the average energy we have to spend to remove a nucleon from the nucleus to infinite distance. It is given by total binding energy divided by the mass number of the nucleus. The BE curve has a broad maximum in the range $$A = 30$$  to $$A = 120$$  corresponding to average binding energy per nucleon $$= 8\,MeV.$$  The peak value of the maximum is $$8\,MeV/N$$  for $$_{26}F{e^{56}}.$$

Work done to stop the $$\alpha $$ particle is equal to K.E.
$$\eqalign{ & \therefore qV = \frac{1}{2}m{v^2} \Rightarrow q \times \frac{{K\left( {Ze} \right)}}{r} = \frac{1}{2}m{v^2} \cr & \Rightarrow r = \frac{{2\left( {2e} \right)K\left( {Ze} \right)}}{{m{v^2}}} = \frac{{4KZ{e^2}}}{{m{v^2}}} \cr & \Rightarrow r \propto \frac{1}{{{v^2}}}{\text{ and }}r \propto \frac{1}{m}. \cr} $$

If $${A_0}$$ is the initial activity of radioactive sample, then activity at any time
$$\eqalign{ & A = {A_0}{e^{ - \lambda t}} \cr & {\text{or}}\,\,1 \times {10^6} = 4 \times {10^6}{e^{ - \lambda \times 20}} \cr & {\text{or}}\,\,{e^{ - 20\lambda }} = \frac{1}{4} \cr} $$
The count rate after 100 hour is given by
$$\eqalign{ & A' = {A_0}{e^{ - \lambda \times 100}} = {A_0}{e^{ - 100\lambda }} = {A_0}{\left[ {{e^{ - 20\lambda }}} \right]^5} \cr & = 4 \times {10^6}{\left[ {\frac{1}{4}} \right]^5} = 3.91 \times {10^3}\,{\text{per}}\,{\text{second}} \cr} $$

The reaction can be written as
$$_1{H^2}{ + _1}{H^2}{ \to _2}H{e^4} + {\text{energy}}$$
The energy released in the reaction is the difference of binding energies of daughter and parent nuclei.
Hence, energy released $$ = {\text{binding energy of}}{\,_2}H{e^4} - 2 \times {\text{binding energy of}}{{\text{ }}_1}{H^2}$$
$$ = 28 - 2 \times 2.2 = 23.6\,MeV$$