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A spherical drop of radius 10^–6 m has absorbed 40 electrons. The energy required to give an additional electron to it is

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a

5.96 × 10²⁴ J

b

5.7 × 10^–21 J

c

9.21 × 10^–23 J

d

9.2 × 10^–21 J

answer is A.

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Detailed Solution

$\begin{array}{l} W = q_{1} V = q_{1} \times \frac{l}{4 {πε}_{0}} \frac{q}{r}; Here q = 40 e \\ w = (1 .6 \times 10^{- 19}) \times \frac{9 \times 10^{9} \times (40 \times 1 .6 \times 10^{- 19})}{10^{- 6}} \\ W = 9 .2 \times 10^{- 21} J \end{array}$

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