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 $2V$, assuming that $40\%$ of the emitted $\beta$-particles escapes the sources? ___ $\left(6.94\times 10^{-4}\right)$

Answer 1

Answer: 6.94

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 $2=\frac{1}{4\pi {ϵ}_0}\frac{\left(2\times 10^{10}\right)t\times \left(1.6\times 10^{-19}\right)}{10^{-2}}$
$\therefore t=6.94\times 10^{-4}s$