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Q.

In case of the hydrogen atom, the energy of the electron in the n^th orbit is given by

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a

$\frac{- 2.18 X 10^{- 18}}{n^{2}} c a l / a t o m$

b

$\frac{- 313.6}{n^{2}} J o u l e / M o l e$

c

$\frac{- 313.6}{n^{2}} K c a l / M o l e$

d

$\frac{- 2.18 X 10^{- 18}}{n^{2}} J / M o l e$

answer is B.

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Detailed Solution

The energy of electron in Kcal / mole from

$E_{n} = \frac{- 2 π^{2} Z^{2} {me}^{4}}{n^{2} h^{2}} (n^{th} orbit)$

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