KEY CONCEPT: 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 risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to 100 ergs of energy absorbed by 1 gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist L. H. Gray); one gray equals 100 rad.

The time required for the number of parent nuclei to fall to $$50\% $$  is called half-life $$T$$ and may be related to disintegration constant $$\lambda $$ as follows.
Since,
$$0.5\,{N_0} = {N_0}{e^{ - \lambda t}}\,\left[ {_{\lambda = \,{\text{decay constant}}}^{N = {\text{ Final No}}{\text{. of nuclei}}\, = 0.5\,{N_0}}} \right]$$
$${N_0} = $$ Initial No. of nuclei
$$\lambda = $$ decay constant
we have, $$\lambda T = {\log _e}2$$
$$\therefore \lambda = \frac{{{{\log }_e}2}}{T}$$

By conservation of momentum, we have
$$\eqalign{ & 0 = {{\vec P}_\alpha } + {{\vec P}_d} \cr & \Rightarrow {{\vec P}_\alpha } = - {{\vec P}_d}\,\,{\text{or}}\,\,{P_\alpha } = {P_d} = P \cr} $$
The kinetic energy released in the process
$$\eqalign{ & K = {K_\alpha } + {K_p} = \frac{{{P^2}}}{{2{m_\alpha }}} + \frac{{{P^2}}}{{2{m_d}}} \cr & = \frac{{{P^2}}}{{2{m_\alpha }}}\left( {1 + \frac{{{m_\alpha }}}{{{m_d}}}} \right) = \frac{{{{\left( {\frac{h}{\lambda }} \right)}^2}}}{{2{m_\alpha }}}\left( {1 + \frac{{{m_\alpha }}}{{{m_d}}}} \right) \cr} $$
After substituting the given values, we get
$$K = 6.25\,MeV$$
If $${{m_p}}$$ is the mass of the parent nucleus, then
$$\eqalign{ & K + \left( {{m_\alpha } + {m_d}} \right){c^2} = {m_p}{c^2} \cr & {\text{or}}\,\,6.25 + \left( {223.61 + 4.002} \right){c^2} = {m_p}{c^2} \cr} $$
After simplifying, we get
$${m_p} = 227.62\,amu.$$

$$M\left( {_8{O^{16}}} \right) = M\left( {_7{N^{15}}} \right) + 1{m_P}$$
binding energy of last proton
$$\eqalign{ & = M\left( {{N^{15}}} \right) + {m_P} - M\left( {_1{O^{16}}} \right) \cr & = 15.00011 + 1.00783 - 15.99492 \cr & = 0.01302\,amu \cr & = 12.13\,MeV \cr} $$

Binding energy of a nucleus containing $$N$$ neutrons and $$Z$$ protons is
$$\eqalign{ & BE = \left[ {N{m_n} + Z{m_p} - m\left( {N,Z} \right)} \right]{c^2} \cr & \Rightarrow \frac{{BE}}{{{c^2}}} = N{m_n} + Z{m_p} - m\left( {N,Z} \right) \cr & \Rightarrow m\left( {N,Z} \right) = N{m_n} + Z{m_p} - \frac{{BE}}{{{c^2}}} \cr} $$

Let $$t$$ = time for the potential of metal sphere to rise by one volt.
Now $$\beta $$-particles emitted in this time $$ = \left( {6.25 \times {{10}^{11}}} \right) \times t$$
Number of $$\beta $$-particles escaped in this time $$ = \left( {\frac{{80}}{{100}}} \right)\left( {6.25 \times {{10}^{10}}} \right)t = 5 \times {10^{10}}t$$
$$\therefore $$ Charge acquired by the sphere in $$t\sec .$$
$$Q = \left( {5 \times {{10}^{10}}t} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right) = 8 \times {10^{ - 19}}t\,.......\left( {\text{i}} \right)$$
($$\because $$ emission of $$\beta $$-particle lends to a charge $$e$$ on metal sphere)
The capacitance $$C$$ of a metal sphere is given by $$C = 4\pi {\varepsilon _0} \times r$$
$$ = \left( {\frac{1}{{9 \times {{10}^9}}}} \right) \times \left( {\frac{{{{10}^{ - 3}}}}{2}} \right) = \frac{{{{10}^{ - 12}}}}{{18}}\,farad\,......\left( {{\text{ii}}} \right)$$
we know that $$Q = C \times V\left\{ {{\text{Here}}\,V = 1\,volt} \right\}$$
$$\therefore \left( {8 \times {{10}^{ - 9}}} \right)t = \left( {\frac{{{{10}^{ - 12}}}}{{18}}} \right) \times 1$$
Solving it for $$t,$$ we get $$t = 6.95\,\mu \sec .$$

The electromagnetic spectrum is as follows

∴ $$\gamma $$ -rays has least wavelength

Mean life of radioactive substance is given by
$$\tau = \frac{1}{\lambda },\,\,\left( {\lambda \,{\text{is}}\,{\text{decay constant}}} \right)$$
Also, it is given that $$t = 2\tau $$
So, $$t = 2 \times \frac{1}{\lambda } = \frac{2}{\lambda }$$
Thus, mass remained after time $$t$$ is
\[M = {M_0}{e^{ - \lambda t}}\,\,\left[ {\begin{array}{*{20}{c}} {M = {\rm{Final\,mass}}}\\ {{M_0} = {\rm{Inital\,mass}}}\\ {\lambda = {\rm{Decay\,constant}}} \end{array}} \right]\]
$$\eqalign{ & = 10{e^{ - \lambda \times \frac{2}{\lambda }}}\,\,\,\left( {\because {M_0} = 10\,g} \right) \cr & = 10{e^{ - 2}} \cr & = \frac{{10}}{{{e^2}}} \cr & = 1.35\,g \cr} $$

$$\eqalign{ & {N_1} = {N_0}{e^{ - 5\lambda t}}\,......\left( {\text{i}} \right) \cr & {N_2} = {N_0}{e^{ - \lambda t}}\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) by Eq. (ii), we obtain
$$\eqalign{ & \frac{{{N_1}}}{{{N_2}}} = \frac{1}{e} = \frac{{{e^{ - 5\lambda t}}}}{{{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}} \cr & \therefore {e^{ - 1}} = {e^{ - 4\lambda t}} \cr & {\text{or}}\,1 = 4\lambda t \cr & {\text{or}}\,\,t = \frac{1}{{4\lambda }} \cr} $$