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Excitation energy of hydrogen atom is 13.6eV. Match the following:
COLUM N-I COLUM N-II
A) Energy of second excited state of hydrogen P) -3.4eV
B) Energy of fourth state of He⁺ Q) -108.9eV
C) Energy of first excited state of Li²⁺ R) -1.5eV
D) Energy absorbed when the electron in Li²⁺ makes transition from 3^rd to 1^st orbit S) -30.6eV

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A-R, B-P, C-S, D-Q

A-P, B-Q, C-S, D-R

A-R, B-Q, C-P, D-S

A-S, B-Q, C-R, D-P

answer is B.

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Detailed Solution

The energy of an electron in a orbit of a hydrogen like ion will be given by $E = - \frac{13.6}{n^{2}} z^{2}$

Energy of second excited state of hydrogen = $E = - \frac{13.6}{3^{2}} 1^{2} = - 1.5 eV$

Energy of fourth state of He⁺= $E = - \frac{13.6}{5^{2}} 2^{2} = - 3.4 eV$

Energy absorbed when the electron in Li²⁺ makes transition from 3^rd to 1^st orbit $∆ E = - (\frac{1}{1^{2}} - \frac{1}{3^{2}}) \times 13.6 \times 3^{2} = - 108.8 eV$

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