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Q.

In a hydrogen atom the electron and a proton are bound together at a separation of about $0.53 A^{0}$ The potential energy of electron proton system is

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a

$- 54.4 ev$

b

$- 27.2 ev$

c

$- 13.6 ev$

d

$0$

answer is B.

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Detailed Solution

$U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} = \frac{9 \times 10^{9} \times 1.6 \times 1.6 \times 10^{- 38}}{0.53 \times 10^{- 10}} = 27.2 ev$

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