Radioactivity MCQ Questions & Answers in Modern Physics | Physics

Let the unknown product nucleus be $$_Z{X^A}.$$
The reaction can be written as
\[\begin{array}{*{20}{c}} {_8{O^{16}}}\\ {\left( {{\rm{Oxygen}}} \right)} \end{array} + \begin{array}{*{20}{c}} {_1{H^2}}\\ {\left( {{\rm{deuterium}}} \right)} \end{array} \to \begin{array}{*{20}{c}} {_Z{X^A}}\\ {\left( {{\rm{unknown}}\,{\rm{nucleus}}} \right)} \end{array} + \begin{array}{*{20}{c}} {_2H{e^4}}\\ {\left( {\alpha - {\rm{particle}}} \right)} \end{array}\]
Conservation of mass number between product and reactant of above reaction gives,
$$16 + 2 = A + 4 \Rightarrow A = 14$$
Conservation of atomic number between reactant and product of above reaction gives
$$8 + 1 = Z + 2 \Rightarrow Z = 7$$
Thus, the unknown product nucleus is nitrogen $$\left( {_7{N^{14}}} \right).$$
NOTE
Fusion reaction can take place at very high temperature $$\left( { \approx {{10}^8}K} \right)$$   and very high pressure which can be provided at sun or by fission of atom bomb.

In $$\gamma $$ -decay, the atomic number and mass number do not change.

Alpha particle is a positive particle. An alpha particle has $$3.2 \times {10^{ - 19}}C$$   charge twice the negative charge of an electron. The mass of an $$\alpha $$-particle is $$6.645 \times {10^{ - 27}}\,kg$$    which is equal to mass of helium nucleus. When two electrons are emitted by a helium atom, a nucleus of helium remains which has charge equal to that of two electrons. Actually alpha $$\left( \alpha \right)$$ particle is a nucleus of helium. Hence, it is also called as doubly-ionised helium atom.

The binding energy for $$_1{H^1}$$ is around zero and also not given in the question so we can ignore it
$$\eqalign{ & Q = 2\left( {4 \times 7.06} \right) - \left( {7 \times 5.60} \right) \cr & = 2\left( {{E_{bn}}{\text{of}}\,He} \right) - \left( {{E_{bn}}{\text{of}}\,Li} \right) \cr & = \left( {8 \times 7.06} \right) - \left( {7 \times 5.60} \right) \cr & = \left( {56.48 - 39.2} \right)MeV \cr & \therefore Q = 17.28\,MeV \simeq 17.3\,MeV \cr} $$

Here, conservation of linear momentum can be applied

$$\eqalign{ & 238 \times 0 = 4u + 234v \Rightarrow \,\therefore v = - \frac{4}{{234}}u \cr & \therefore {\text{speed}} = \left| {\vec v} \right| = \frac{4}{{234}}u \cr} $$

$$\eqalign{ & \frac{{dN}}{{dt}} = - N\left( {{\lambda _1} + {\lambda _2}} \right) = - N\lambda \cr & \Rightarrow \lambda = {\lambda _1} + {\lambda _2} \cr & {\text{so}}\,\frac{1}{t} = \frac{1}{{{t_1}}} + \frac{1}{{{t_2}}}\left[ {{\text{as}}\,\lambda = \frac{{0.693}}{t}} \right] \cr} $$

$$\eqalign{ & N = {N_0}{e^{ - \frac{\lambda }{t}}} \cr & = {10^6}{e^{ - \left[ {\frac{{\ln 2}}{{20}} \times 10} \right]}} \cr & = {10^6}{e^{ - \ln \sqrt 2 }} \cr & = \frac{{{{10}^6}}}{{\sqrt 2 }} \cr & = 7.07 \times {10^5} \cr & \approx 7 \times {10^5} \cr} $$

When a nucleus emits an alpha particle, its mass number decreases by 4 and charge/atomic no. decreases by 2. In $$\beta $$-particle emission, mass remains same but atomic number is increased by one, In $$\gamma $$-decay, daughter nucleus has the same charge number and same mass number as those of parent nucleus. Hence, sequence is

We know that $$\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} = {\left( {\frac{1}{2}} \right)^{\frac{t}{{{T_{\frac{1}{2}}}}}}}$$
$$\eqalign{ & \Rightarrow \frac{1}{{1 + 15}} = \frac{1}{{16}} = {\left( {\frac{1}{2}} \right)^4} = {\left( {\frac{1}{2}} \right)^{\frac{t}{{50}}}} \cr & t = 4 \times 50 \cr & t = 200\,yr \cr} $$

According to law of radioactivity
\[\begin{array}{l} \frac{N}{{{N_0}}} = {e^{ - \lambda t}}\,\,......\left( {\rm{i}} \right)\\ \Rightarrow \frac{{{N_0}}}{N} = {e^{\lambda t}}\,\,\left[ {\begin{array}{*{20}{c}} {N = {\rm{final}}\,{\rm{concentration}}}\\ {{N_0} = {\rm{initial\,concentration}}}\\ {\lambda = {\rm{decay\,constant}}} \end{array}} \right] \end{array}\]
Taking logarithm on both sides of Eq. (i), we have
$$\eqalign{ & {\log _e}\left( {\frac{{{N_0}}}{N}} \right) = {\log _e}\left( {{e^{\lambda t}}} \right) \cr & = \lambda t\,{\log _e}e = \lambda t \cr} $$
As we know that, $${\log _e}x = 2.3026\,{\log _{10}}x$$
Making substitution, we get
$$\lambda = \frac{{2.3026\,{{\log }_{10}}\left( {\frac{{9750}}{{975}}} \right)}}{5}\,\,\left[ {\because {N_0} = 9750\,{\text{counts}}/\min \,{\text{and}}\,N = 975\,{\text{counts}}/\min } \right]$$
$$ = \frac{{2.3026}}{5}{\log _{10}}10 = \frac{{2.3026}}{5}{\min ^{ - 1}} = 0.461\,{\min ^{ - 1}}$$