Beta rays are same as cathode rays as both are stream of electrons.

Let initially there are total $${N_0}$$ number of nuclei
At time $$t$$ $$\frac{{{N_B}}}{{{N_A}}} = 0.3\left( {{\text{given}}} \right)$$
$$\eqalign{ & \Rightarrow {N_B} = 0.3\,{N_A} \cr & {N_0} = {N_A} + {N_B} = {N_A} + 0.3\,{N_A} \cr & \therefore {N_A} = \frac{{{N_0}}}{{1.3}} \cr} $$
As we know $${N_t} = {N_0}{e^{ - \lambda t}}$$
$$\eqalign{ & {\text{or,}}\,\,\frac{{{N_0}}}{{1.3}} = {N_0}{e^{ - \lambda t}} \cr & \frac{1}{{1.3}} = {e^{ - \lambda t}} \cr & \Rightarrow \ln \left( {1.3} \right) = \lambda t \cr & {\text{or,}}\,\,t = \frac{{\ln \left( {1.3} \right)}}{\lambda } \cr & \Rightarrow t = \frac{{\ln \left( {1.3} \right)}}{{\frac{{\ln \left( 2 \right)}}{T}}} = \frac{{\ln \left( {1.3} \right)}}{{\ln \left( 2 \right)}}T \cr} $$

Given, $$_1{H^2}{ + _1}{H^2}{ \to _2}{H^4} + Q$$
The total binding energy of the deutrons $$ = 4 \times 1.15 = 4.60\,MeV$$
The total binding energy of alpha particle $$ = 4 \times 7.1 = 28.4\,MeV$$
The energy released in the process $$ = 28.4 - 4.60 = 23.8\,MeV.$$

Alpha particles are positively charged particles with charge $$+2e$$  and mass $$4\,m.$$  Emission of an $$\alpha $$-particle reduces the mass of the radionuclide by 4 and its atomic number by 2. $$\beta $$-particles are negatively charged particles with rest mass as well as charge same as that of electrons. $$\gamma $$-particles carry no charge and mass.
Radioactive transition will be as follows
$$\eqalign{ & _Z^AX \to _{Z + 1}^AY + \beta _{ - 1}^0 \cr & _{Z + 1}^AY \to _{Z - 1}^{A - 4}\beta + \alpha _2^4 \cr & _{Z + 1}^{A - 4}\beta \to _{Z - 1}^{A - 4}\beta + \gamma _0^0 \cr} $$

Energy of $$\gamma $$ -ray photon = Rest mass energy + $$K.E.$$
$$ = 2\left( {0.5} \right) + 0.78 = 1.78\,MeV$$

Suppose an initial radio nuclide $$I$$ decays to a final product $$F$$ with a half - life $${T_{\frac{1}{2}}}.$$
At any time, $${N_I} = {N_0}{e^{ - \lambda t}}$$
Number of product nuclei $$ = {N_F} = {N_0} - {N_I}$$
$$\eqalign{ & \frac{{{N_F}}}{{{N_I}}} = \frac{{{N_0} - {N_I}}}{{{N_I}}} = \left( {\frac{{{N_0}}}{{{N_I}}} - I} \right) \cr & \frac{{{N_0}}}{{{N_I}}} = \left( {1 + \frac{{{N_F}}}{{{N_I}}}} \right) = 1 + 0.5 = 1.5 \cr & \therefore \frac{{{T_{\frac{1}{2}}}\ln \left( {1.5} \right)}}{{\ln 2}} = 4.5 \times {10^9}\frac{{\ln \left( {\frac{3}{2}} \right)}}{{\ln 2}}{\text{year}} \cr} $$

$$\eqalign{ & \frac{{{N_1}}}{{{N_2}}} = {e^{\left( {{\lambda _2} - {\lambda _1}} \right)}} = \frac{1}{e} \cr & \Rightarrow t = \frac{1}{{{\lambda _2} - {\lambda _1}}} = 2 \cr} $$

$$Q = 4\left( {{x_2} - {x_1}} \right)$$

Using
$$\eqalign{ & \frac{M}{{{M_0}}} = {\left( {\frac{1}{2}} \right)^{\frac{t}{T}}} = {\left( {\frac{1}{2}} \right)^{\frac{{120}}{{24}}}} = {\left( {\frac{1}{2}} \right)^5} = \frac{1}{{32}} \cr & {\text{So}}\,\,M = \frac{{{M_0}}}{{32}} = \frac{{0.1\,mg}}{{32}} = 3.125\,\mu g. \cr} $$

The decay rate $$R$$ of a radioactive material is the number of decays per second.
From radioactive decay law, $$ - \frac{{dN}}{{dt}} \propto N$$
$${\text{or}}\,\, - \frac{{dN}}{{dt}} = \lambda N$$
i.e. Rate of reaction is directly proportional to the initial concentration of reactants.
$$\eqalign{ & {\text{Thus,}}\,\,R = - \frac{{dN}}{{dt}}\,\,{\text{or}}\,\,R \propto N \cr & {\text{or}}\,\,R = \lambda N\,\,{\text{or}}\,\,R = \lambda {N_0}{e^{ - \lambda t}}\,......\left( {\text{i}} \right) \cr} $$
where $${R_0} = \lambda {N_0}$$   is the activity of the radioactive material at time $$t = 0.$$
$$\eqalign{ & {\text{At}}\,{\text{time}}\,{t_1},\,\,{R_1} = {R_0}{e^{ - \lambda {t_1}}}\,......\left( {{\text{ii}}} \right) \cr & {\text{At}}\,{\text{time}}\,{t_2},\,\,{R_2} = {R_0}{e^{ - \lambda {t_2}}}\,......\left( {{\text{iii}}} \right) \cr} $$
Dividing Eq. (ii) by Eq. (iii), we have
$$\eqalign{ & \frac{{{R_1}}}{{{R_2}}} = \frac{{{e^{ - \lambda {t_1}}}}}{{{e^{ - \lambda {t_2}}}}} = {e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} \cr & {\text{or}}\,\,{R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}} \cr} $$