A radioactive source in the form of metal sphere of diameter $${10^{ - 3}}m$$  emits beta particle at a constant rate of $$6.25 \times {10^{10}}$$   particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that $$80\% $$  of the emitted beta particles escape from the source?

Solution :
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 $$t = 6.95\,\mu \sec .$$