Question

A radioactive source in the form of a metal sphere of radius ${10}^{-2}$ m emits beta particles (electrons) at the
rate of $5\times {10}^{10}$ particles per second. The source is electrically insulated.
How long will it take for its potential to be raised by
2 V assuming that 40% of the emitted beta particles escape
from the source.

Answer 1

Beta particles (i.e., electrons) escaped in t seconds.

$\eta =\frac{40}{100}\times 5\times {10}^{10}t=2\times {10}^{10}t$

Therefore, the deficiency of electrons from a conductor results in positive charge. As 40% beta particles escape from the source, positive charge gained by source (metal sphere),

$q=ne=2\times {10}^{10}t\times 1.6\times {10}^{-19}C=3.2\times {10}^{-9}tC$

Due to this, charge, the potential acquired by sphere is

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

Substituting given values, we get

$2=9.0\times {10}^9\times \frac{3.2\times {10}^{-9}t}{{10}^{-2}}ort=\frac{1}{1440}s\simeq 700\mu s$