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Energy of an electron in n^thorbit of hydrogen atom is

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$\frac{- n^{2} h^{2}}{2 π^{2} {kme}^{4}}$

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

$\frac{- 2 π^{2} Z^{2} {me}^{4}}{n^{2} h^{2}}$

$\frac{- n^{2} h^{2}}{4 π^{2} {kme}^{4}}$

answer is A.

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Detailed Solution

According to the formula to calculate the energy of electron in any orbit.

$\frac{- 2 π^{2} Z^{2} {me}^{4}}{n^{2} h^{2}}$

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