Radioactivity MCQ Questions & Answers in Modern Physics | Physics

$$\eqalign{ & {\text{If}}\,m = 1\,u,c = 3 \times {10^8}m{s^{ - 1}},\,{\text{then}} \cr & E = 931\,MeV\,{\text{i}}{\text{.e}}{\text{.}}\,1\,u = 931\,MeV \cr} $$
Binding energy $$ = 0.042 \times 931 = 39.10\,MeV$$
∴ Binding energy per nucleon $$ = \frac{{39.10}}{7} = 5.58 \approx 5.6\,MeV$$

Ratio of number of half life taken is given as :
After 16 days
$$\eqalign{ & {T_{A\frac{1}{2}}} = \frac{{16}}{4} = 4; \cr & {T_{B\frac{1}{2}}} = \frac{{16}}{8} = 2; \cr & N = {N_0}{\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow \frac{{{N_A}}}{{{N_B}}} = \frac{1}{{{2^4}}}:\frac{1}{{{2^2}}} = {2^2}:{2^4} \cr & = 4:16, = 1:4 \cr} $$

\[X\xrightarrow{{{\lambda _1}}}Y\xrightarrow{{{\lambda _2}}}Z\]
At radioactive equilibrium,
$$\eqalign{ & \left( {{\lambda _X}} \right) \times \left( {{N_X}} \right) = \left( {{\lambda _Y}} \right) \times \left( {{N_Y}} \right) \cr & \frac{{{\lambda _X}}}{{{\lambda _Y}}} = \frac{{{N_Y}}}{{{N_X}}}\frac{{{{\left( {{t_{\frac{1}{2}}}} \right)}_Y}}}{{{{\left( {{t_{\frac{1}{2}}}} \right)}_X}}} = \frac{{{N_Y}}}{{{N_X}}}\,\,{\text{or}}\,\,\frac{{{{\left( {{t_{\frac{1}{2}}}} \right)}_Y}}}{{1.6 \times {{10}^{10}}}} \cr & = \frac{1}{{3.2 \times {{10}^9}}}{\left( {{t_{\frac{1}{2}}}} \right)_Y} = 5\,{\text{years}} \cr} $$

$$\eqalign{ & A = {A_0}{\left( {\frac{1}{2}} \right)^n};n = {\text{number of half lives}}{\text{.}} \cr & \frac{{{A_0}}}{{16}} = {A_0}{\left( {\frac{1}{2}} \right)^n}\,\,\,\,\,\therefore {\left( {\frac{1}{2}} \right)^4} = {\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow n = 4 \cr & \therefore t = \left( {4 \times 100} \right)\mu {\text{s}} = 400\mu s \cr} $$

Ratio of $$X:Y$$  is given $$ = 1:7$$
$$\frac{{{m_x}}}{{{m_y}}} = \frac{1}{7} \Rightarrow 7{m_x} = {m_y}$$
Let the initial total mass be $$m.$$
$$\eqalign{ & \Rightarrow {m_x} + {m_y} = m \Rightarrow \frac{{{m_y}}}{7} + {m_y} = m \cr & \Rightarrow \frac{{8{m_y}}}{7} = m \Rightarrow {m_y} = \frac{7}{8}m \cr} $$
only $$\frac{1}{8}$$ part remains

So, time taken to become $$\frac{1}{8}$$ unstable part
$$ = 3 \times {T_{\frac{1}{2}}} = 3 \times 1.4 \times {10^9} = 4.2 \times {10^9}y$$
Alternative
$$_{{\text{Active}}}^X \to _{{\text{Stable}}}^Y$$
As we know that
$$\eqalign{ & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & \frac{1}{{1 + 7}} = \frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} \cr & \Rightarrow n = 3 \cr & {\text{As,}}\,\,{T_{\frac{1}{2}}} = \frac{t}{n} \cr & \therefore t = {T_{\frac{1}{2}}} \times n \cr & = 1.4 \times {10^9} \times 3 = 4.2 \times {10^9}y \cr} $$

$$\eqalign{ & {N_1}{\lambda _1} = {N_2}{\lambda _2}\frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{2}{3} \cr & \Rightarrow \frac{{{N_2}}}{{{N_1}}} = \frac{2}{3} \cr} $$
⇒ ratio of no. of nuclei of $$X$$ to that of $$Y = \frac{3}{2}$$

For this substance $$7$$ days correspond to $$3.5$$ half - lives. Over $$3$$ half- lifes the sample reduces to $$\frac{1}{{{2^3}}} = \frac{1}{8}$$  of its initial mass. After 4 half- lifes, the sample has only $$\frac{1}{{{2^4}}} = \frac{1}{{16}}$$  of its initial mass. Hence, after $$3.5$$ half - lives the sample must contain somewhere between $$\frac{1}{8}$$ and $$\frac{1}{16}$$ of its initial mass. Hence $$5\,g$$  is somewhere between $$\frac{1}{8}$$ and $$\frac{1}{16}$$ of the initial mass.
So, the initial mass is somewhere between
$$8 \times 5 = 40\,g\,\,{\text{and}}\,\,16 \times 5 = 80\,g.$$

Number of half-lives
$$n = \frac{t}{{{T_{\frac{1}{2}}}}} = \frac{{18}}{3} = 6\,\,\left[ {{T_{\frac{1}{2}}} = {\text{half life period}}} \right]$$
Amount remained after $$n$$ half-lives
$$N = {N_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{N = {\text{ final count}}}^{{N_0} = \,\,{\text{initial count}}}} \right]$$
$$\eqalign{ & {\text{Given,}}\,\,{N_0} = 300\,g \cr & \therefore N = 300{\left( {\frac{1}{2}} \right)^6} = 300 \times \frac{1}{{64}} = 4.68\,g \cr} $$
Alternative
Total time of decay given
$$\eqalign{ & t = \frac{{2.303}}{\lambda }{\log _{10}}\left( {\frac{{300}}{N}} \right) \cr & {\text{but}}\,\,\lambda = \frac{{0.693}}{T} = \frac{{0.693}}{3} = 0.231/h \cr & \therefore t = \frac{{2.303}}{{0.231}}{\log _{10}}\left( {\frac{{300}}{N}} \right) \cr & {\text{Given}}\,\,t = 18\;h \cr & {\text{So,}}\,\,18 = \frac{{2.303}}{{0.231}}{\log _{10}}\left( {\frac{{300}}{N}} \right) \cr & {\text{or}}\,\,{\log _{10}}\left( {\frac{{300}}{N}} \right) = \frac{{0.231}}{{2.303}} \times 18 \cr & {\text{or}}\,\,\frac{{300}}{N} = {\left( {10} \right)^{18}}\,\,{\text{or}}\,\,N = \frac{{300}}{{{{\left( {10} \right)}^{18}}}} = 4.68\,g \cr} $$

$$\eqalign{ & {T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda } \cr & \therefore \lambda = \frac{{\ln 2}}{{{T_{\frac{1}{2}}}}} \cr & \Rightarrow {\lambda _A} = \frac{{\ln 2}}{{{T_A}}},{\lambda _B} = \frac{{\ln 2}}{{{T_B}}} \cr & \Rightarrow \frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{T_B}}}{{{T_A}}}. \cr} $$

Given, for $$^{14}C$$
\[\begin{array}{l} {A_0} = 16\,dis\,{\min ^{ - 1}}\,{g^{ - 1}}\\ A = 12\,dis\,{\min ^{ - 1}}\,{g^{ - 1}} \end{array}\]
$$\eqalign{ & {t_{\frac{1}{2}}} = 5760\,{\text{years}} \cr & {\text{Now,}}\,\lambda = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} = \frac{{0.693}}{{5760}}\,{\text{per}}\,{\text{year}} \cr} $$
Then, from, $$t = \frac{{2.303}}{\lambda }{\log _{10}}\frac{{{A_0}}}{A}$$
$$ = \frac{{2.303 \times 5760}}{{0.693}}{\log _{10}}\frac{{16}}{{12}} \approx 2391\,{\text{years}}$$