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Q.

The correct expression derived for the energy of an electron in the n^th energy level is for H-atom:

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a

$E_{n} = - \frac{2 π^{2} {me}^{4} K^{2}}{n^{2} h^{2}}$

b

$E_{n} = - \frac{π^{2} {me}^{4} K^{2}}{2 n^{2} h^{2}}$

c

$E_{n} = \frac{2 π^{2} {me}^{4} K^{2}}{n^{2} h^{2}}$

d

$E_{n} = - \frac{2 π^{2} {me}^{2} K^{2}}{n^{2} h^{2}}$

answer is D.

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Detailed Solution

$E_{n} = - \frac{2 π^{2} {me}^{4} K^{2}}{n^{2} h^{2}}$ this is energy of nth electron.

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