Question

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

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

Answer 1

Answer: (A)

For $H:\frac{1}{\lambda }=R_H\times 1^2\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right).....$(1)
$\frac{1}{{\lambda }_{H{e}^{+}}}=R_H\times 2^2\times \left(\frac{1}{4}-\frac{1}{9}\right).....$(2)
From (1) & (2) $\frac{{\lambda }_{H{e}^{+}}}{\lambda }=\frac{9}{5}$
${\lambda }_{H{e}^{+}}=\lambda \times \frac{9}{5}$
${\lambda }_{H{e}^{+}}-\frac{9\lambda }{5}$