A spherical drop of radius $10$ has absorbed $40$ electrons. The energy required to given an additional electron to it is.

9.2 \times 10^{- 21} J

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5.7 \times 10^{- 21} J

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9.21 \times 10^{- 23} J

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4 K e^{2}

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Solution

The correct option is D $4 K e^{2}$
We know that,
Energy $=$ Potential $\times$ Charge
$E = (\frac{k q}{r}) \times e$
$E = \frac{k (40 e) (e)}{10}$
$E = 4 k {(e)}^{2}$
$E n e r g y = 4 {k e}^{2}$

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