The energy required to take the electron from second orbit to third orbit of hydrogen atom is approximately

48.5 e V

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10.2 e V

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1.9 e V

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13.6 e V

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Solution

The correct option is C $1.9 e V$
The energy of an electron in $n^{t h}$ energy level is
= $- 13.6 \times \frac{Z^{2}}{n^{2}}$

Here, $Z = 1$ for hydrogen atom

Hence $E_{2}$ = $\frac{- 13.6}{2^{2}}$ = $- 3.4$ eV
and $E_{3}$ = $\frac{- 13.6}{3^{2}}$ = $- 1.51$ eV

Hence energy required for electron to be taken from the second orbit to the third orbit
= $E_{3} - E_{2}$
= $- 1.51 - (- 3.4) e V$
= $1.89 e V$
which is approximately $1.9 e V$ .

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