A radioactive source, in the form of a metallic sphere of radius `10^(-2)m` emits `beta-`particles at the rate of `5xx10^(10)` particles per second. The source is electrically insulated. How long will it take for its potential to be raised by `2V`, assuming that `40%` of the emitted `beta-`particles escape the source.

Answer

`700 mu s`

Explanation

Beta particles (i.e., electrons) escaped in t seconds.
`eta = (40)/(100) xx 5 xx 10^10 t = 2 xx 10^10 t`
Therefore, the deficeiency 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 = n e = 2 xx 10^10 t xx 1.6 xx 10^-19 C = 3.2 xx 10^-9 t C`
Due to this charge, the potential acquired by sphere is
`V = (1)/(4 pi epsilon_0) q/( R)`
Substituting given values, we get
`2 = 9.0 xx 10^9 xx (3.2 xx 10^-9 t)/(10^-2)` or `t = (1)/(1440) s ~= 700 mu s`.