Question

The shortest wavelength of H atom in the Lyman series is ${\lambda }_1$. The longest wavelength in the Balmer series of $H{e}^{+}$ is

• A. $\frac{5{\lambda }_1}{9}$
• B. $\frac{36{\lambda }_1}{5}$
• C. $\frac{27{\lambda }_1}{5}$
• D. $\frac{9{\lambda }_1}{5}$

Answer 1

The correct option is D $\frac{9{\lambda }_1}{5}$

Shortest wavelength $\to \text{Max energy}(\infty \to 1)$ (Lyman series)

$\frac{1}{{\lambda }_1}=R_H(1{)}^2[\frac{1}{1}-0]$

$\frac{1}{{\lambda }_1}=R_H\Rightarrow R_H=\frac{1}{{\lambda }_1}$

For Balmer series,
$\frac{1}{\lambda }=R_H(2{)}^2[\frac{1}{2^2}-\frac{1}{3^2}]\Rightarrow R_H(4)\left(\frac{9-4}{36}\right)$

$\frac{1}{\lambda }=\frac{5R_H}{9}\Rightarrow \lambda =\frac{9}{5R_H}=\frac{9{\lambda }_1}{5}$
Hence the correct answer is option (d).