Question

A radioactive source, in the form of a metallic sphere of radius $10^{-2} m$ emits $\beta$-particles at the rate of $5 \times 10^{10}$ particles per second. The source is electrically insulated. How long (in second) will it take for its potential to be raised by $2 V$, assuming that $40 \%$ of the emitted $\beta$-particles escapes the sources? ___ $\left(6.94 \times 10^{-4}\right)$

Answer 1

In time $t$, the total number of $\beta$-particles emitted $=5 \times 10^{10} t$.
As only $40 \%$ escape from the surface, so $N =0.40 \times\left(5 \times 10^{10}\right) t=2 \times 10^{10} t$.
The charge develops due to escape of each $\beta$ - particle is $1.6 \times 10^{-19} C$, so total charge escapes in time $t, q=\left(2 \times 10^{10}\right) t \times\left(1.6 \times 10^{-19}\right) C$. Thus we can write,
$V =\frac{1}{4 \pi \in_0} \frac{q}{R} $
or $ 2 =\frac{1}{4 \pi \epsilon_0} \frac{\left(2 \times 10^{10}\right) t \times\left(1.6 \times 10^{-19}\right)}{10^{-2}} $
$ \therefore t =6.94 \times 10^{-4} s $