Radioactivity MCQ Questions & Answers in Modern Physics | Physics

The number of $$\alpha $$ -particles released = 8
Therefore the atomic number should decrease by 16
The number of $${\beta ^ - }$$ -particles released = 4
Therefore the atomic number should increase by 4.
Also the number of $${\beta ^ + }$$ particles released is 2, which should decrease the atomic number by 2.
Therefore the final atomic number is $$92 - 16 + 4 - 2 = 78$$

Number of atoms left after $$n$$ half-lives is given by
$$\eqalign{ & N = {N_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{N = {\text{final count rate of the }}n{\text{ half life}}}^{{N_0} = \,\,{\text{initial count}}}} \right] \cr & {\text{or}}\,\,\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & {\text{where}}\,\,n = \frac{t}{{{T_{\frac{1}{2}}}}} \cr & \therefore n = \frac{{6400}}{{1600}} = 4\,\,\left[ {_{{T_{\frac{1}{2}}} = 1600}^{t\, = \,6400}} \right] \cr & \therefore \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^4} = \frac{1}{{16}} \cr} $$

$$\lambda = \frac{{2.303}}{{280}}\log \frac{{{A_0}}}{{6000}} = \frac{{2.303}}{{420}}\log \frac{{{A_0}}}{{3000}}$$
On solving, we get, $${A_0} = 24000\,dps.$$

$$\frac{7}{8}$$ of $$Cu$$  decays in 15 minutes.
$$\therefore $$ $$Cu$$  undecayed $$ = N = 1 - \frac{7}{8} = \frac{1}{8} = {\left( {\frac{1}{2}} \right)^3}$$
$$\therefore $$ No. of half lifes = 3
$$n = \frac{t}{T}\,\,{\text{or}}\,\,3 = \frac{{15}}{T}$$
$$ \Rightarrow $$ $$T =$$  half life period $$ = \frac{{15}}{3} = 5\,{\text{minutes}}$$

$$\left| {\frac{{\frac{d}{N}}}{{dt}}} \right| = \lambda N$$
Number of radius nuclei in $$mg = \frac{{{N_A}m}}{{226}}$$
Decay constant,
$$\eqalign{ & \lambda = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}\frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}} \cr & \left| {\frac{{dN}}{{dt}}} \right| = 10 = \frac{{6.02 \times {{10}^{23}}m}}{{226}} \times \frac{{0.693}}{{1620 \times 3.16 \times {{10}^7}}} \cr & \therefore m = \frac{{10 \times 226 \times 162 \times 3.16 \times {{10}^7}}}{{6.02 \times {{10}^{23}} \times 0.693}} \cr & = 2.77 \times {10^{ - 10}}g \cr & = 2.77 \times {10^{ - 13}}kg \cr} $$

We know that the rate of integration $$\left| { - \frac{{dN}}{{dt}}} \right| = A$$
$$\eqalign{ & \therefore A = {A_0}{e^{ - \lambda t}}\,\,{\text{or}}\,\,2700 = 4750{e^{ - \lambda \times 5}} \cr & {\text{or}}\,\,\lambda = 0.1131\,{\text{per}}\,{\text{minute}} \cr & {\text{Half}}\,{\text{life}}\,\,{t_{\frac{1}{2}}} = \frac{{0.693}}{\lambda } = \frac{{0.693}}{{0.1131}} = 6.1\,{\text{minute}} \cr} $$

$$\eqalign{ & \frac{{^{14}C}}{{^{12}C}} = 1.3 \times {10^{ - 12}} \cr & 12{\text{ }}g{\text{ contain}}\,6.022 \times {10^{23}}\,{\text{atoms}} \cr & {\text{No}}{\text{.}}\,{\text{atoms}}\,{\text{of}}{\,^{14}}C = 6.022 \times {10^{23}} \times 1.3 \times {10^{ - 12}} \cr & = 7.8286 \times {10^{11}} \cr & \Rightarrow N = {N_0}{e^{ - \lambda t}} - \frac{{dN}}{{dt}} = {N_0}{e^{ - \lambda t}} \times \lambda \cr & \Rightarrow - \frac{{dN}}{{dt}} = N \times \lambda \cr & \frac{{180}}{{60}} = 7.8286 \times {10^{11}} \times \lambda \cr & \Rightarrow \frac{1}{\lambda } = \frac{{7.8286 \times {{10}^{11}}}}{3} \cr & \Rightarrow \lambda = 0.3832 \times {10^{ - 11}} \cr & {t_{\frac{1}{2}}} = \frac{{0.691}}{\lambda } = \frac{{0.691}}{{0.3832 \times {{10}^{ - 11}}}} = 1.80 \times {10^{11}}\sec \cr & {\text{Half life}} = 1.8032 \times {10^{11}}\sec = 0.5740 \times {10^4} \cr & {\text{year = 5740 years}}{\text{.}} \cr} $$

Let $$E$$ be the energy of proton, then
$$\eqalign{ & E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right] \cr & \Rightarrow E = 56.48 - 39.2 = 17.28\,MeV \cr} $$

$$\frac{7}{8}$$ of $$Cu$$ decays in 15 minutes.
∴ $$Cu$$ undecayed $$ = N = 1 - \frac{7}{8} = \frac{1}{8} = {\left( {\frac{1}{2}} \right)^3}$$
∴ No. of half lifes = 3
$$n = \frac{t}{T}\,{\text{or}}\,3 = \frac{{15}}{T}$$
$$ \Rightarrow T = {\text{half}}\,{\text{life}}\,{\text{period}} = \frac{{15}}{3} = 5\,{\text{minutes}}$$

Let $$Z$$ be the charge number and $$A$$ be the mass number of particle $$X,$$ then conservation of charge number gives
$$6 = 5 + 1 + Z \Rightarrow Z = 0$$
Conservation of mass number gives,
$$11 = 11 + 0 + A \Rightarrow A = 0$$
$$X$$ is a particle of zero charge and zero mass. This particle may be neutrino or antineutrino. As we know that for positive $$\beta $$-particle, neutrino is emitted and with negative $$\beta $$-particle, antineutrino is emitted.
Thus, in this case neutrino will be emitted.