For $$A + B \to C + \varepsilon ,\varepsilon $$    is positive. This is because $${E_b}$$ for $$C$$ is greater than the $${E_b}$$ for $$A$$ and $$B.$$
Again for $$F \to D + E + \varepsilon ,\varepsilon $$    is positive. This is because $${E_b}$$ for $$D$$ and $$E$$ is greater than $${E_b}$$ for $$F.$$

In two half lives, the activity will remain $$\frac{1}{4}$$ of its initial activity.

KEY CONCEPT:
Distance of closest approach
$$\eqalign{ & {r_0} = \frac{{Ze\left( {2e} \right)}}{{4\pi {\varepsilon _0}E}} \cr & {\text{Energy, }}E = 5 \times {10^6} \times 1.6 \times {10^{ - 19}}J \cr & \therefore {r_0} = \frac{{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} \cr & \Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm \cr} $$

The activity of a radioactive substance is
\[R = {R_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {\begin{array}{*{20}{c}} {R = {\rm{Final\,number}}}\\ {{R_0} = {\rm{Initial\,number}}} \end{array}} \right]\]
Here, $$n$$ = number of half-lives
$$\eqalign{ & = \frac{t}{{{T_{\frac{1}{2}}}}} = \frac{{24}}{6} = 4\,\,\left[ {{T_{\frac{1}{2}}} = {\text{Half}}\,\,{\text{life}}\,\,{\text{period}}} \right] \cr & {\text{and}}\,\,R = 0.01\,\mu C \cr & {\text{So,}}\,\,0.01 = {R_0}{\left( {\frac{1}{2}} \right)^4} \cr & {\text{or}}\,\,{R_0} = 0.01 \times {\left( 2 \right)^4} \cr & = 0.01 \times 16 \cr & = 0.16\,\mu C \cr} $$

$$\eqalign{ & N = {N_0}{e^{ - \lambda t}}q = {q_0}{e^{\frac{{ - t}}{{RC}}}}\,{\text{and}}\,\,\frac{N}{q} = {\text{constant}} \cr & {\text{when}}\,{e^{ - \lambda t}} = {e^{\frac{t}{{RC}}}} \cr & \Rightarrow \lambda = \frac{1}{{RC}}R = 200\,\Omega . \cr} $$

The $$\alpha $$-particle can be represented as $$_2H{e^4}$$  and $$\beta $$-particle as $$_{ - 1}{\beta ^0}.$$  So, after emission of one $$\alpha $$-particle the mass number of resultant nucleus decreases by 4 unit and atomic number by 2 unit. Similarly, after emission of one $$\beta $$-particle the atomic number increases by 1 unit keeping its mass number same. So, according to reaction (assuming $$_Z{X^A}$$  the initial nucleus)
$$\eqalign{ & _Z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}\,\,\left( {\alpha - {\text{particle}}} \right) \cr & {\text{and}}\,{\,_{Z - 2}}{Y^{A - 4}}{ \to _Z}{X^{A - 4}} + 2\left( {_{ - 1}{\beta ^0}} \right)\,\,\left( {2\beta - {\text{particle}}} \right) \cr} $$
So, by one $$\alpha $$ and two $$\beta $$-emissions the atomic number remains unchanged. i.e. formation of isotopes takes place.

The binding energy curve has a broad maximum in the range $$A = 30$$  to $$A = 120$$  corresponding to average binding energy per nucleon $$= 8\,MeV.$$   The peak value of the maximum is $$8.8\,MeV/N$$   for $$_{26}F{e^{56}}.$$

$$\eqalign{ & N = {N_0}{e^{ - \lambda t}} \cr & \Rightarrow {N_y} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right) \cr} $$
Rate of formation of $$Y = \frac{{dN}}{{dt}} = + \lambda {N_0}{e^{ - \lambda t}}$$
$$\eqalign{ & X \to Y \cr & {\text{At}}\,\,t = 0,R = \lambda {N_0} \cr & \Rightarrow t = 0\,{N_0} \cr & t = \infty ,R = 0 \cr & \Rightarrow t = t\,N \cr & {N_y} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right) \cr} $$

From equation (Ist) there is 1 $$ \propto $$-decay in which $$B$$ has atomic no. 2 less than $$A.$$ In IInd case there is 2-$$\beta $$-decay in which $$C$$ has atomic no. 2 greater than $$B,$$ Since $$A$$ and $$C$$ have same atomic no. so they are called isotopes.

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