Radioactivity MCQ Questions & Answers in Modern Physics | Physics

$$\beta $$ -rays are fast moving beam of electrons.

$$Q$$ value of the reaction
$$\eqalign{ & Q = \left( {2 \times 4 \times 7.06 - 7 \times 5.6} \right) = 17.28\,MeV\,.......\left( {\text{i}} \right) \cr & {K_P} + Q = 2{K_\alpha }\,.......\left( {{\text{ii}}} \right) \cr & \sqrt {2{m_P}{K_P}} = 2\sqrt {2{m_\alpha }{K_\alpha }} \cos \alpha \cr & {K_P} = {K_\alpha }\,.......\left( {{\text{iii}}} \right) \cr & {\text{So}}\,\,{K_P} = 17.28\,MeV. \cr} $$

By conservation of momentum, $${p_1} = {p_2}$$
$$\eqalign{ & \sqrt {2{K_1}{m_1}} = \sqrt {2{K_2}{m_2}} \cr & \Rightarrow \sqrt {2{K_1}\left( {216} \right)} = \sqrt {2{K_2}\left( 4 \right)} \cr & \Rightarrow {K_2} = 54{K_1}\,......\left( {\text{i}} \right) \cr & {\text{Also,}}\,{K_1} + {K_2} = 5.5MeV\,......\left( {{\text{ii}}} \right) \cr} $$
Solve equation (i) and (ii)

Let $${N_0}$$ be the initial number of nuclei, then
$${N_1} = {N_0}{e^{ - \lambda {t_1}}}\,\,{\text{and}}\,\,{N_2} = {N_0}{e^{ - \lambda {t_2}}}$$
∴ Number of nuclei decayed $$ = {N_1} - {N_2}$$
$$\eqalign{ & = {N_0}\left( {{e^{ - \lambda {t_1}}} - {e^{ - \lambda {t_2}}}} \right) = \frac{{{A_0}}}{\lambda }\left( {{e^{ - \lambda {t_1}}} - {e^{ - \lambda {t_2}}}} \right) \cr & = \frac{{{A_1} - {A_2}}}{\lambda } = \left( {{A_1} - {A_2}} \right)\tau . \cr} $$

The range of energy of $$\beta $$-particles is from zero to some maximum value.

$$\eqalign{ & 4_2^4He \to _8^{16}O \cr & B.E. = \Delta m \times 931.5\,MeV \cr & = \left( {4 \times 4.0026 - 15.9994} \right) \times 931.5 = 10.24\,MeV \cr} $$

The total number of protons bombarded
$$ = \frac{{it}}{\lambda } = \frac{{{{10}^{ - 4}} \times 3600}}{{1.6 \times {{10}^{ - 19}}}} = 22.5 \times {10^{17}}$$
Number of $$^7Be$$  produced
$$N = \frac{{22.5 \times {{10}^{17}}}}{{1000}} = 22.5 \times {10^{14}}$$
We know that activity
$$\eqalign{ & A = \lambda N\,\,{\text{or}}\,\,A = \left( {\frac{{0.693}}{{{t_{\frac{1}{2}}}}}} \right)N \cr & \therefore {t_{\frac{1}{2}}} = 0.693\frac{N}{A} = 0.693 \times \frac{{22.5 \times {{10}^{14}}}}{{1.8 \times {{10}^8}}} = 8.63 \times {10^6}s \cr} $$

The radioactive substances emit $$\alpha $$ -particles (Helium nucleus), $$\beta $$ -particles (electrons) and neutrinoes.

KEY CONCEPT: Intensity $$I = {I_0}.{e^{ - \mu d}}$$
Applying logarithm on both sides,
$$\eqalign{ & - \mu d = \log \left( {\frac{I}{{{I_0}}}} \right) \cr & - \mu \times 36 = \log \left( {\frac{{\frac{I}{8}}}{I}} \right)\,......\left( {\text{i}} \right) \cr & - \mu \times d = \log \left( {\frac{{\frac{I}{2}}}{I}} \right)\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\frac{{36}}{d} = \frac{{\log \left( {\frac{1}{8}} \right)}}{{\log \left( {\frac{1}{2}} \right)}} = \frac{{3\log \left( {\frac{1}{2}} \right)}}{{\log \left( {\frac{1}{2}} \right)}} = 3\,{\text{or}}\,d = \frac{{36}}{3} = 12mm$$

For a nucleus to disintegrate in two half life, the probability is $$\frac{3}{4}$$ as $$75\% $$ of the nuclei will disintegrate in this time.