Radioactivity MCQ Questions & Answers in Modern Physics | Physics

Energy released $$ = \left( {80 \times 7 + 120 \times 8 - 200 \times 6.5} \right)$$
$$ = 220\,MeV.$$

$$\eqalign{ & \frac{{{C^{14}}}}{{{C^{12}}}} = \frac{1}{4} = {\left( {\frac{1}{2}} \right)^{\frac{t}{{5700}}}} \cr & \Rightarrow \frac{t}{{5700}} = 2 \cr & \Rightarrow t = 11400\,{\text{years}} \cr} $$

Total time given $$= {80\,\min }$$
Number of half-lives of $$A,$$ $${n_A} = \frac{{80\,\min }}{{20\;\,\min }} = 4$$
Number of half-lives of $$B,$$ $${n_B} = \frac{{80\,\min }}{{40\;\,\min }} = 2$$
Number of nuclei remained undecayed $$N = {N_0}{\left( {\frac{1}{2}} \right)^n}$$
where $${N_0}$$ is initial number of nuclei and $$N$$ is final number of nuclei
So for two different cases $$\left( A \right)$$  and $$\left( B \right),$$
$$\eqalign{ & \frac{{{N_A}}}{{{N_B}}} = \frac{{{{\left( {\frac{1}{2}} \right)}^{{n_A}}}}}{{{{\left( {\frac{1}{2}} \right)}^{{n_B}}}}}\,\,{\text{or}}\,\,\frac{{{N_A}}}{{{N_B}}} = \frac{{{{\left( {\frac{1}{2}} \right)}^4}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = \frac{{\left( {\frac{1}{{16}}} \right)}}{{\left( {\frac{1}{4}} \right)}} \cr & {\text{or}}\,\,\frac{{{N_A}}}{{{N_B}}} = \frac{1}{4} \cr} $$

The reaction can be shown as

Thus, the resulting nucleus is the isotope of parent nucleus and is $$_n{X^{m - 4}}.$$

Number of undecayed atom after time $${t_2};$$
$$\frac{{{N_0}}}{3} = {N_0}{e^{ - \lambda {t_2}}}\,......\left( {\text{i}} \right)$$
Number of undecayed atom after time $${t_1};$$
$$\eqalign{ & \frac{{2{N_0}}}{3} = {N_0}{e^{ - \lambda {t_1}}}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right),{e^{ - \lambda {t_2}}} = \frac{1}{3} \cr & \Rightarrow - \lambda {t_2} = {\log _e}\left( {\frac{1}{3}} \right)\,......\left( {{\text{iii}}} \right) \cr & {\text{From}}\,\left( {{\text{ii}}} \right) - {e^{ - \lambda {t_2}}} = \frac{2}{3} \cr & \Rightarrow - \lambda {t_1} = {\log _e}\left( {\frac{2}{3}} \right)\,......\left( {{\text{iv}}} \right) \cr} $$
Solving (iii) and (iv), we get
$${t_2} - {t_1} = 20min$$

There is no change in the proton number and the neutron number as the $$\gamma $$-emission takes place as a result of excitation or de-excitation of nuclei. $$\gamma $$-rays have no charge or mass.

Let the radioactive substance be $$_Z^AX.$$
Radioactive transition is given by
\[_Z^AX\xrightarrow{{ - \alpha }}_{Z - 2}^{A - 4}X\xrightarrow{{ - 2\beta }}_Z^{A - 4}X\]
The atoms of element having same atomic number but different mass numbers are called isotopes.
So, $$_Z^AX$$  and $$_Z^{A - 4}X$$  are isotopes.

$$\eqalign{ & \frac{{ - dN}}{{dt}} = {\lambda _1}N + {\lambda _2}N \cr & \Rightarrow {\log _e}\frac{N}{{{N_o}}} = - \left( {{\lambda _1} + {\lambda _2}} \right)t \cr} $$
when $${N_o}$$ is initial number of atoms
$$\eqalign{ & {\text{Here }}{\lambda _1} = \frac{{0.693}}{{1620}}\,{\text{and}}\,{\lambda _2} = \frac{{0.693}}{{810}}; \cr & \frac{N}{{{N_o}}} = \frac{1}{4} \Rightarrow {\log _e}\frac{1}{4} = - \left( {\frac{{0.693}}{{1620}} + \frac{{0.693}}{{810}}} \right)t \cr & \Rightarrow t = 1080{\text{ years}} \cr} $$

Fraction remains after $$n$$ half lives
$$\eqalign{ & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} = {\left( {\frac{1}{2}} \right)^{\frac{t}{T}}} \cr & {\text{Given}}\,\,N = \frac{{{N_0}}}{e} \Rightarrow \frac{{{N_0}}}{{e{N_0}}} = {\left( {\frac{1}{2}} \right)^{\frac{5}{T}}} \cr & {\text{or}}\,\,\frac{1}{e} = {\left( {\frac{1}{2}} \right)^{\frac{5}{T}}} \cr} $$
Taking log on both sides, we get
$$\eqalign{ & \log 1 - \log e = \frac{5}{T}\log \frac{1}{2} \cr & \Rightarrow - 1 = \frac{5}{T}\left( { - \log 2} \right) \cr & \Rightarrow T = 5{\log _e}2 \cr} $$
Now, let $$t’$$ be the time after which activity reduces to half
$$\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^{\frac{{t'}}{{5{{\log }_e}2}}}} \Rightarrow t' = 5{\log _e}2$$

Momentum of a photon $$p = \frac{{h\nu }}{c}$$
Hence, recoil energy, $$E = \frac{{{p^2}}}{{2M}}$$
$$\therefore E = \frac{{{{\left( {\frac{{h\nu }}{c}} \right)}^2}}}{{2M}}\,{\text{or}}\,\,E = \frac{{{h^2}{\nu ^2}}}{{2M{c^2}}}$$