Question

A radioactive source in the form of a metallic sphere of radius ${10}^{-2}\text{m}$ emits $\beta -$ particles at the rate of $5\times {10}^{10}$ particles per second. The source is electrically insulated. How long $($in $\mu \text{s})$ will it take for its potential to be raised by $2\text{V}?$

Assume that $40\%$ of the emitted $\beta -$ particles escape from the source.

Round off to the nearest integer.

Answer 1

Let $t$ be the time required to raise the potential by $2\text{V}$.

The number of $\beta -$ particles emitted in $t$ second will be,

$n=(5\times {10}^{10})t$

Number of $\beta -$ particles escaping from sphere is $40\%$.

Therefore, positive charge developed on the sphere in times $t$ is,

$Q=+(5\times {10}^{10})t\times \left(0.40\right)e$

$\Rightarrow Q=+(2\times {10}^{10})t\times 1.6\times {10}^{-19}\text{C}$

$\Rightarrow Q=(3.2\times {10}^{-9})t\text{C}$

The potential at the surface of the charged metallic sphere is given by,

$V=\frac{Q}{4\pi {ϵ}_{0}R}$

$\Rightarrow 2=\frac{3.2\times {10}^{-9}\times t}{4\pi {ϵ}_0\times \left({10}^{-2}\right)}=\frac{(9\times {10}^9)\times (3.2\times {10}^{-9})t}{{10}^{-2}}$

$\Rightarrow 2=(9\times 100\times 3.2)t$

$\Rightarrow t=\frac{2}{2880}=694.44\times {10}^{-6}\text{s}$

$\therefore t=694.44\mu \text{s}\approx 694\mu \text{s}$