Question

The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is $20,397c{m}^{-1}$. The wave number of the energy for the same transition in $H{e}^{+}$ is

• A. $5,099c{m}^{-1}$
• B. $20,497c{m}^{-1}$
• C. $40,994c{m}^{-1}$
• D. $81,588c{m}^{-1}$

Answer 1

The correct option is D $81,588c{m}^{-1}$

Using $\frac{1}{\lambda }=\overline{v}=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\Rightarrow \overline{v}\infty Z^2\Rightarrow \frac{\overline{v_2}}{\overline{v_1}}={\left(\frac{Z_2}{Z_1}\right)}^2={\left(\frac{Z}{1}\right)}^2=4\Rightarrow \overline{v_2}=\overline{v}\times 4=81588c{m}^{-1}$