First slide

Hydrogen -like ion

Question

The wave number of energy emitted when electron jumps from fourth orbit to second orbit in hydrogen in 20,497 cm^-1. The wave number of energy for the same transition in He⁺ is

Easy

A

5,099 cm^-1
B

20,497 cm^-1
C

40,994 cm^-1
D

81,988 cm^-1

Solution

$\bar{v} = \frac{1}{λ} = {Rz}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{{n_{2}}^{2}}); ∴ \bar{v} α z^{2}$
$∴ \frac{{\bar{v}}_{1}}{{\bar{v}}_{2}} = {(\frac{z_{1}}{z_{2}})}^{2}; ∴ \frac{20497}{{\bar{v}}_{2}} = {(\frac{1}{2})}^{2}$
$∴ {\bar{v}}_{2} = 81988$

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