When a parent nucleus emits a $$\beta $$-particle (i.e, an electron) mass number remains same because mass of electron is negligibly low. Atomic number is increased by one. The nucleus $$_{48}C{d^{115}}$$  after two successive $$\beta $$-decays will give $$_{50}S{n^{115}}.$$

Energy is released when stability increases. This will happen when binding energy per nucleon increases.

Reactant	Product
Reaction (a) $$60 \times 8.5MeV = 510MeV$$	$$2 \times 30 \times 5 = 300MeV$$
Reaction (b) $$120 \times 7.5 = 900MeV$$	$$\left( {90 \times 8 + 30 \times 5} \right) = 870MeV$$
Reaction (c) $$120 \times 7.5 = 900MeV$$	$$2 \times 60 \times 8.5 = 1020MeV$$
Reaction (d) $$90 \times 8 = 720MeV$$	$$\left( {60 \times 8.5 + 30 \times 5} \right) = 600MeV$$

The penetrating power is dependent on velocity.
For a given energy, the velocity of $$\gamma $$ radiation is highest and $$\alpha $$-particle is least.

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

The equation for initial and final count rate is
$$\eqalign{ & N = {N_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{{N_0} = {\text{ initial rate of radio - active atom}}}^{N = \,\,{\text{final count rate}}}} \right] \cr & {\text{where,}}\,n = \frac{t}{{{T_{\frac{1}{2}}}}} \cr & {\text{Here,}}\,\,n = \frac{{120}}{{30}} = 4\,\,\left[ {\because t = 2h = 2 \times 60\min = 120\min } \right] \cr & \therefore \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^4} = \frac{1}{{16}} \cr & {\text{or}}\,\,{N_0} = 16 \times N = 16 \times 5 = 80\,{s^{ - 1}} \cr} $$

Number of nuclei remained after time $$t$$ can be written as
$$N = {N_0}{e^{ - \lambda t}}$$
where, $${N_0}$$  is initial number of nuclei of both the substances.
$$\eqalign{ & {N_1} = {N_0}{e^{ - 5\lambda t}}\,.......\left( {\text{i}} \right) \cr & {\text{and}}\,\,{N_2} = {N_0}{e^{ - \lambda t}}\,.......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) by Eq. (ii), we obtain
$$\frac{{{N_1}}}{{{N_2}}} = {e^{\left( { - 5\lambda + \lambda } \right)t}} = {e^{ - 4\lambda t}} = \frac{1}{{{e^{4\lambda t}}}}$$
But, we have given
$$\frac{{{N_1}}}{{{N_2}}} = {\left( {\frac{1}{e}} \right)^2} = \frac{1}{{{e^2}}}$$
Hence, $$\frac{1}{{{e^2}}} = \frac{1}{{{e^{4\lambda t}}}}$$
Comparing the powers, we get
$$2 = 4\lambda t\,\,{\text{or}}\,\,t = \frac{2}{{4\lambda }} = \frac{1}{{2\lambda }}$$

Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to Uranium which has a big nuclei and less B.E/nucleon.
In other words, Iodine and Yttrium are more stable and therefore possess less energy and less rest mass. Also when Uranium nuclei explodes, it will convert into $$I$$ and $$Y$$ nuclei having kinetic energies.

Activity of sample becomes $$2500$$  from $$5000$$  in $$75$$ days therefore its half life is $$75$$ days, so
$$\eqalign{ & R = \frac{{{R_0}}}{{{2^{\frac{{150}}{{75}}}}}} = 5000 \cr & \Rightarrow {R_0} = 5000 \times 4 = 20,000 \cr} $$

The wavelength of spectrum is given by
$$\frac{1}{\lambda } = {Rz^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)\,\,\,\,\,\,{\text{where}}\,R = \frac{{1.097 \times {{10}^7}}}{{1 + \frac{m}{M}}}$$
where $$m$$ = mass of electron
$$M$$ = mass of nucleus.
For different $$M, R$$  is different and therefore $$\lambda $$ is different

Effective half life is calculated as
$$\eqalign{ & \frac{1}{T} = \frac{1}{{{T_1}}} + \frac{1}{{{T_2}}} \cr & \frac{1}{T} = \frac{1}{{16}} + \frac{1}{{48}} \cr & \Rightarrow T = 12\,{\text{years}} \cr} $$
Time in which $$\frac{3}{4}$$ will decay is 2 half lives = 24 years