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The Ionisation potential of Hydrogen is 2.17× 10^–11 erg/atom. The energy of the electron
in the second orbit of the hydrogen atom is

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$- \frac{2.17 \times 10^{- 11}}{2}$

$- \frac{2.17 \times 10^{- 11}}{2^{2}}$

$- \frac{2.17 \times 10^{- 17}}{2^{2}}$

$- \frac{2.17 \times 10^{11}}{2^{2}}$

answer is B.

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Detailed Solution

$E = - \frac{2.17 \times 10^{- 11}}{n^{2}} e r g . a t o m^{- 1} s e c o n d o r b i t i . e n = 2 E = - \frac{2.17 \times 10^{- 11}}{2^{2}} e r g . a t o m^{- 1}$

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