The energy of electron in the $n^{t h}$ Bohr orbit of H-atom is :

\frac{- 13.6}{n^{2}} e V

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\frac{- 13.6}{n} e V

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\frac{- 13.6}{n^{4}} e V

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\frac{- 13.6}{n^{3}} e V

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Solution

The correct option is D $\frac{- 13.6}{n^{2}} e V$
Energy of electron in $n^{t h}$ Bohr orbit is given by the formula:

$E = \frac{- 13.6}{n^{2}} e V$

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The energy of an electron in the first Bohr orbit of H - atom is - 13.6 eV. Then, which of the following statement(s) is/are correct for $H e^{+}$ ?

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The energy of electron in the nth Bohr orbit of H-atom is :

The correct option is D −13.6n2eVEnergy of electron in nth Bohr orbit is given by the formula:E=−13.6n2 eV

The energy of electron in the $n^{t h}$ Bohr orbit of H-atom is :

The correct option is D $\frac{- 13.6}{n^{2}} e V$
Energy of electron in $n^{t h}$ Bohr orbit is given by the formula:

$E = \frac{- 13.6}{n^{2}} e V$