A radioavtive source in the form of a metal sphere of daimeter `10^(-3)` m emits `beta`-particles at a constant rate of `6.25 xx 10^(10)` particles per second. If the source is electrically insulated, how long will it take for its potential to rise by `1.0 V`, assuming that `80%` of the emitted `beta`-particles escape the socurce?

Answer

`6.95 mu s`

Explanation

Let t be the time for potential of metal sphere to rise by `1.0 V`. Then, up to this `beta`-particles emitted from sphere are
`N =(6.25 xx 10^(10)) xxt`
Number of `beta`-particles escaped in this time are
`N_(e) =(80)/(100)xx(6.25 xx 10^(10))t`
`=5 xx10^(10)t`
Thus, charge acquired by the speher in t s
`Q=(5 xx10^(10)t) xx(1.6 xx10^(-19))`
`=8xx10^(-9)` t columb
(emission of `beta`- particles leads to a charege e on the metal sphere) The capacitance `C` of a metal sphere is given by
`C=4pi epsilon_(0) xx r`
`=((1)/(9xx10^(9)))xx((10^(-3))/(2))=(10^(12))/(18)farad`
We know that `Q=C xx V" "(here, V=1" volt")`
or `( 8xx 10^(-9))t=((10^(-12))/(10))xx1`
Solving it for `t`, get `t =6.95 mu s` .