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Physics Question on atom structure models

The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is 20,397 cm−120,397\text{ }c{{m}^{-1}} . The wavelength of energy for the same transition in He+ is :

A

81,988cm−181,988c{{m}^{-1}}

B

40,994cm−140,994c{{m}^{-1}}

C

20,497cm−120,497c{{m}^{-1}}

D

5,099 cm−15,099\text{ }c{{m}^{-1}}

Answer

5,099 cm−15,099\text{ }c{{m}^{-1}}

Explanation

Solution

As λ∝1Z2\lambda \propto \frac{1}{{{Z}^{2}}} Hence wavelength of energy for same transition in He is, here Z = 2 hence, 203974=5099 cm−1\frac{20397}{4}=5099\text{ }c{{m}^{-1}}