Radioactivity MCQ Questions & Answers in Modern Physics | Physics

From conservation of momentum $${m_1}{v_1} = {m_2}{v_2}$$
$$\eqalign{ & \Rightarrow \left( {\frac{{{m_1}}}{{{m_2}}}} \right) = \left( {\frac{{{v_2}}}{{{v_1}}}} \right){\text{given}}\,\frac{{{v_1}}}{{{v_2}}} = 2 \cr & \Rightarrow \frac{{{m_1}}}{{{m_2}}} = \frac{1}{2} \Rightarrow \frac{{r_1^3}}{{r_2^3}} = \frac{1}{2} \Rightarrow \left( {\frac{{{r_1}}}{{{r_2}}}} \right) = {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}} \cr} $$

$$\eqalign{ & B.E. = \Delta m{c^2} = \Delta \times 931\,MeV \cr & = \left[ {2\left( {1.0087 + 1.0073} \right) - 4.0015} \right] \times 931 \cr & = 28.4\,MeV \cr} $$

$$\eqalign{ & A - 12 - \left( {Z - 8} \right) \cr & \therefore {\text{Required}}\,{\text{ratio}}\, = \frac{{A - Z - Y}}{{Z - 8}} \cr} $$

The total energy released
$$E$$ = Energy released in fission process + energy released in $$\alpha $$-decay process
$$\eqalign{ & = {N_F} \times 200 + {N_\alpha } \times \left( {0.005517 \times 931} \right) \cr & = \left( {\frac{8}{{100}} \times {{10}^{20}}} \right) \times 200 + \left( {\frac{{92}}{{100}} \times {{10}^{20}}} \right) \times \left( {0.005517 \times 931} \right) \cr & = 20.725 \times {10^{20}}MeV \cr} $$
Power output $$P = \frac{E}{t}$$
$$\eqalign{ & = \frac{{20.725 \times {{10}^{20}} \times 1.6 \times {{10}^{ - 13}}}}{{{{10}^{13}}}} \cr & = 3.3 \times {10^{ - 5}}W. \cr} $$

After every half-life, the mass of the substance reduces to half its initial value.

Use the relation $$N = {N_0}{e^{ - \lambda t}}$$   where the decay constant $$\lambda = \left( {\frac{{0.693}}{{{T_{{\text{half}}}}}}} \right).$$    The half life period $${{T_{{\text{half}}}}}$$  is given to be $$10\,\min.$$  The amount $$N$$ of radioactive sample remaining at the end of $$5\,\min$$  comes out to be about $$0.707\,g,$$  so that the amount of sample decayed is $$\left( {1 - 0.707} \right) = 0.293\,g.$$

Number of half-lives $$n = \frac{t}{{{T_{\frac{1}{2}}}}} = \frac{{{\text{30}}\,{\text{days}}}}{{{\text{10}}\,{\text{days}}}} = 3\,\,\left[ {{T_{\frac{1}{2}}} = {\text{half life period}}} \right]$$
So, number of undecayed radioactive nuclei is given by
$$\eqalign{ & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{{N_0} = \,{\text{Initial number}}}^{N = \,{\text{Final}}\,{\text{number}}}} \right] \cr & {\text{or}}\,\,N = {N_0}{\left( {\frac{1}{2}} \right)^n} = 4 \times {10^{10}}{\left( {\frac{1}{2}} \right)^3} \cr & = 4 \times {10^{10}} \times \frac{1}{8} \cr & = 0.5 \times {10^{10}} \cr} $$
Thus, number of nuclei decayed after 30 days
$$\eqalign{ & = {N_0} - N \cr & = 4 \times {10^{10}} - 0.5 \times {10^{10}} \cr & = 3.5 \times {10^{10}} \cr} $$

In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to disappearance of a fraction of the total mass. If the quantity of mass disappearing is $$\Delta m,$$  then the binding energy is
$$BE = \Delta m{c^2}$$
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the consituent neutrons and protons. We can then write.
$$\Delta m = Z{m_p} + N{m_n} - m\left( {A,Z} \right)$$
where $$m\left( {A,Z} \right)$$  is the mass of the atom of mass number $$A$$ and atomic number $$Z.$$ Hence, the binding energy of the nucleus is
$$\eqalign{ & BE = \left[ {Z{m_p} + N{m_n} - m\left( {A,Z} \right)} \right]{c^2} \cr & BE = \left[ {Z{m_p} + \left( {A - Z} \right){m_n} - m\left( {A,Z} \right)} \right]{c^2} \cr} $$
where, $$N = A - Z =$$   number of neutrons.

$$N = {N_0}{e^{ - \lambda t}}$$
Here, $$t = 5$$  minutes
$$\eqalign{ & \frac{{{N_0}}}{e} = {N_0}.{e^{ - 5\lambda }} \cr & \Rightarrow 5\lambda = 1,\lambda = \frac{1}{5}, \cr & {\text{Now,}}\,{T_{\frac{1}{2}}} = \frac{{\ell n2}}{\lambda } = 5\,\ell n2 \cr} $$

Charged particles are deflected in magnetic field.