The energy of an electron present in Bohr’s second orbit of hydrogen atom is

- 1312 J a t o m^{- 1}

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- 328 k J m o l^{- 1}

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- 328 J m o l^{- 1}

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- 164 k J m o l^{- 1}

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Solution

The correct option is B $- 328 k J m o l^{- 1}$
$E_{n} = \frac{- 1312}{n^{2}} k J m o l e^{- 1} = \frac{- 1312}{2^{2}} = - 328 k J m o l^{- 1}$

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Similar questions

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is:
( $a_{0}$ is bohr radius)

a_{o}

a_{0}

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