According to Bohr's theory, the electronic energy of H-atom in Bohr's orbit is given by:

E_{n} = - \frac{2.18 \times 10^{- 19} \times Z}{2 n^{2}} J

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E_{n} = - \frac{2.179 \times 10^{- 18} \times Z^{2}}{n^{2}} J

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E_{n} = - \frac{21.79 \times 10^{- 18} \times Z}{2 n^{2}} J

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E_{n} = - \frac{21.8 \times 10^{- 21} \times Z^{2}}{n^{2}} J

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Solution

The correct option is A $E_{n} = - \frac{2.179 \times 10^{- 18} \times Z^{2}}{n^{2}} J$
According to Bohr's theory, the energy of electron in $n^{t h}$ Bohr's orbit of H-atom or hydrogen-like atoms/ions is given by:
$E_{n} = - R_{H} \frac{Z^{2}}{n^{2}}$
where $R_{H} = 2.18 \times 10^{- 18} J$ and $Z =$ atomic number, $n = n^{t h}$ orbit
Thus, $E_{n} = - 2.18 \times 10^{- 18} \times \frac{Z^{2}}{n^{2}} J$

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