Radioactivity MCQ Questions & Answers in Modern Physics | Physics

For $$40\,g$$  amount,
\[40\,g\xrightarrow[{{\text{half - life}}}]{{20\,s}}20\,g\xrightarrow{{20\,s}}10\,g\]
For $$160\,g$$  amount,
\[\begin{gathered} 160\,g\xrightarrow{{10\,s}}80\,g\xrightarrow{{10\,s}}40\,g \hfill \\ \xrightarrow{{10\,s}}20\,g\xrightarrow{{10\,s}}10\,g \hfill \\ \end{gathered} \]
So, after $$40\,s$$  $${A_1}$$ and $${A_2}$$ will become equal.

$$\beta $$-particles are charged particles emitted by the nucleus.

One point charge is $$\left( {_{92}^{235}U} \right)$$   uranium nucleus
$$\therefore {q_1} = 92e$$

The other point charge is $$\alpha $$ particle $$\therefore {q_2} = + 2e$$
Here the loss in $$K.E.$$  = Gain in $$PE.$$ (till $$\alpha $$-particle reaches the distance $$d$$ )
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = k\frac{{{q_1}{q_2}}}{r} \Rightarrow r = k\frac{{2{q_1}{q_2}}}{{\frac{1}{2}m{v^2}}} \cr & \therefore r = \frac{{9 \times {{10}^9} \times 2 \times 1.6 \times {{10}^{ - 19}} \times 92 \times 1.6 \times {{10}^{ - 19}}}}{{5 \times 1.6 \times {{10}^{ - 13}}}} \cr & = 529.92 \times {10^{ - 16}}m \cr & = 529.92 \times {10^{ - 14}}cm = 5.2992 \times {10^{ - 12}}cm \cr} $$

at time $$t = 0;{A_0} = \frac{{d{N_0}}}{{dt}}$$
$$\eqalign{ & \frac{A}{{{A_0}}} = \frac{{\frac{{dN}}{{dt}}}}{{\frac{{d{N_0}}}{{dt}}}} = \frac{{3260}}{{6520}} \cr & {\text{or}}\,\,\lambda = \frac{{2.303}}{t}\log \frac{{{A_0}}}{A} = \frac{{2.303}}{{2 \times 60}}\log 2 \cr & = 5.78\,{\text{per}}\,\sec \cr} $$

Because radioactivity is a spontaneous phenomenon.

$$\eqalign{ & \frac{A}{{{A_0}}} = \frac{1}{{{2^n}}} \cr & \therefore {2^n} = \frac{{{A_0}}}{{A}} = \frac{{64}}{1} = {2^6} \Rightarrow n = 6 \cr & \therefore {\text{time}} = 6 \times {t_{\frac{1}{2}}} = 6 \times 18 = 108 \cr} $$

Given, Binding energy of $$\left( {_1^2H + _1^3H} \right) = a + b$$
Binding energy of $$_2^4He = c$$
In a nuclear reaction, the resultant nucleus is more stable than the reactants. Hence, binding energy of $$_2^4He$$  will be more than that of $$\left( {_1^2H + _1^3H} \right).$$
Thus, energy released per nucleon = resultant binding energy
Binding energy of product - Binding energy of reactants $$ = c - \left( {a + b} \right) = c - a - b$$

$$\eqalign{ & \frac{{dN}}{{dt}} = \alpha - \lambda N \cr & \int_0^N {\frac{{dN}}{{\alpha - \lambda N}}} = \int_0^t {dt} \cr & {\text{Solving}}\,\,N = \frac{a}{\lambda }\left( {1 - {e^{ - \lambda t}}} \right). \cr} $$

Beta decay involves the emission of either electrons or positrons. The electrons or positrons emitted in a $$\beta $$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative $$\beta $$-decay, a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $$\beta $$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

$$\lambda = \frac{1}{t}{\log _e}\frac{{{A_0}}}{A} = \frac{1}{5}{\log _e}\frac{{5000}}{{1250}} = 0.4{\log _e}2$$