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Energy of an electron in n^th Bohr orbit is given as

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

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

$- \frac{2 {πZe}^{2}}{nh}$

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

answer is B.

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Detailed Solution

Energy of the electron is $\frac{- 2 π^{2} {mZ}^{2} e^{4}}{n^{2} h^{2}}$ where m is the mass of electron, Z is the atomic number, n is the number of orbit, e is the charge of an electron, and h is the Planck's constant.

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