Radioactivity MCQ Questions & Answers in Modern Physics | Physics

According to Einstein, mass-energy equivalence is represented by $$E = m{c^2}.$$
Taking, mass, $$m = 1\,amu = 1.66 \times {10^{ - 27}}kg,$$       and velocity of light in vacuum, $$c = 3.0 \times {10^8}m/s$$
We get, $$E = \left( {1.66 \times {{10}^{ - 27}}} \right) \times {\left( {3 \times {{10}^8}} \right)^2}J$$
$$\eqalign{ & = 1.49 \times {10^{ - 10}}J = \frac{{1.49 \times {{10}^{ - 10}}}}{{1.6 \times {{10}^{ - 13}}}}MeV\,\,\left( {\because 1\,MeV = 1.6 \times {{10}^{ - 13}}J} \right) \cr & = 931.25\,MeV \cr} $$
Hence, $$1\,amu \simeq 931\,MeV$$

Amount of substance remained is
$$\eqalign{ & M = {M_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{{M_0} = \,{\text{initial amount}}}^{M = {\text{ substance remained}}}} \right] \cr & {\text{Given,}}\,\,{M_0} = 100\,g,M = 25\,g, \cr} $$
Half-life of radioactive substance $${T_{\frac{1}{2}}} = 1600\,yr$$
$$\eqalign{ & {\text{So,}}\,\,25 = 100{\left( {\frac{1}{2}} \right)^n} \cr & {\text{or}}\,\,\frac{{25}}{{100}} = {\left( {\frac{1}{2}} \right)^n}{\text{or }}{\left( {\frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^n} \cr} $$
Comparing the power, we have
$$\eqalign{ & n = 2 \cr & {\text{or}}\,\,\frac{t}{{{T_{\frac{1}{2}}}}} = 2 \cr & {\text{or}}\,\,t = 2{T_{\frac{1}{2}}} = 2 \times 1600 \cr & = 3200\,yr \cr} $$

Experimental measurements show that volume of a nucleus is proportional to its mass number $$A.$$ If $$R$$ is the radius of the nucleus assumed to be spherical, then its volume and mass no. relation is given by
$$\eqalign{ & {\text{Volume}}\left( V \right) \propto {\text{mass}}\,{\text{no}}{\text{.}}\left( A \right) \cr & {\text{or}}\,\,\left( {\frac{4}{3}\pi {R^3}} \right) \propto A \cr & {\text{or}}\,\,R \propto {A^{\frac{1}{3}}} \cr & \therefore \frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{{\left( {27} \right)}^{\frac{1}{3}}}}}{{{{\left( {125} \right)}^{\frac{1}{3}}}}} = \frac{3}{5} = \frac{6}{{10}} \cr} $$

Mass of defect
$$\eqalign{ & \Delta m = \left[ {Z{m_P} + \left( {A - Z} \right){m_n}} \right] - M\left( {{U^{238}}} \right) \cr & = \left[ {92 \times 1.007825\left( {238 - 92} \right) \times 1.008665} \right] - 238.050783 \cr & \Delta m = 1.93421\,amu \cr & BE = 1.93421 \times 931.5\,MeV = 1801.7\,MeV \cr} $$

$$_3^7Li + _1^1p \to _4^8Be + _0^0\gamma $$

Note : In a nucleus neutron converts into proton as follows
$$n \to {p^ + } + {e^{ - 1}}$$
Thus, decay of neutron is responsible for $$\beta $$-radiation is carbon atom. origination

$$\eqalign{ & \frac{{dN}}{{dt}} = 50 - \frac{N}{{0.5}} \cr & \int\limits_0^N {\frac{{dN}}{{50 - 2N}}} = \int\limits_0^t {dt} \cr & N = \left( {100\left( {1 - {e^{\frac{{ - t}}{2}}}} \right)} \right) = 25 \cr & t = 2\ln \left( {\frac{4}{3}} \right) \cr} $$

$$\eqalign{ & N = {N_0}{\left( {\frac{1}{2}} \right)^n} \cr & {\text{so,}}\,{\text{for}}\,1\% \,{\text{decay}} \cr & \frac{1}{{100}}{N_0} = {N_0}{\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow \frac{1}{{100}} = {\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow n = \frac{2}{{\log 2}} \cr & \Rightarrow \frac{t}{T} = \frac{2}{{\log 2}} \cr & \Rightarrow t = 6.6T \cr & = 6.6 \times 100 = 660\,{\text{years}} \cr} $$
If $$\alpha $$ and $$\beta $$ are emitted simultaneously.

The mass of radioactive substance remained is,
$$M = {M_0}{\left( {\frac{1}{2}} \right)^n}$$
Here, final mass, $$M = 1\,g,$$   initial mass, $${M_0} = 256\,g,$$   half life period, $${T_{\frac{1}{2}}} = 12.5\,h$$
So, $$1 = 256{\left( {\frac{1}{2}} \right)^n}\,\,{\text{or}}\,\,\frac{1}{{256}} = {\left( {\frac{1}{2}} \right)^n}$$
or $${\left( {\frac{1}{2}} \right)^8} = {\left( {\frac{1}{2}} \right)^n}$$
Comparing the powers on both the sides, we get
$$n = 8 = \frac{t}{{{T_{\frac{1}{2}}}}}$$
$$\eqalign{ & \therefore t = 8{T_{\frac{1}{2}}} \cr & = 8 \times 12.5 \cr & = 100\,h \cr} $$