An isolated radioactive source in the form a metal sphere of radius 1 mm undergoes $β$ –ray at a constant rate of 6.25 × 10¹⁰ /sec. Assuming 70% of the emitted $β$ –particles escape from the source, time after which potential of the sphere rises by 1 V is

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15.5 $μ$ S

20 $μ$ S

25 $μ$ S

10 $μ$ S

answer is A.

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Detailed Solution

$\begin{array}{l} V = \frac{1}{4 {πε}_{0}} \frac{q}{r} \Rightarrow 1 = \frac{9 \times 10^{9} \times n^{1} e}{10^{- 3}} \\ \Rightarrow n^{1} = \frac{10^{- 3}}{9 \times 10^{9} \times 1 .6 \times 10^{- 19}} = \frac{10^{7}}{14 .4} \end{array}$
Since
$\begin{array}{l} n^{1} = \frac{70}{100} (n) [n \to no; of β - particle emitted] \\ \Rightarrow n = \frac{100}{70} \times \frac{10^{7}}{14 .4} \Rightarrow n = \frac{10^{8}}{7 \times 14 .4} \end{array}$
For 6.25 x 10¹⁰ particles → 1sec
For $\frac{10^{8}}{7 \times 14 .4} particles \to' t' \sec$
$∴ t = \frac{(\frac{10^{8}}{14 .4 \times 7})}{6 .25 \times 10^{10}} = 15 .5 \times 10^{- 6} \sec = 15 .5 μsec$