Question

The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is 20,397 cm^–1. The wave number of the energy for the same transition in $He^{+}$ is

• A. 5,099 cm^–1
• B. 20,497 cm^–1
• C. 40,994 cm^–1
• D. 81,998 cm^–1

Answer 1

Answer: (D)

81,998 cm^–1

Answer 2

Using $\frac{1}{\lambda} = \bar{\nu} = RZ^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$ ⇒ $\bar{\nu} \propto Z^{2}$ ⇒

$\frac{{\bar{\nu}}_{2}}{{\bar{\nu}}_{1}} = \left( \frac{Z_{2}}{Z_{1}} \right)^{2} = \left( \frac{Z}{1} \right)^{2} = 4$ ⇒ ${\bar{\nu}}_{2} = \bar{\nu} \times 4 = 81588cm^{- 1}$.