Question

The shortest wavelength of $H{e}^{+}$ ion in Balmer series is x, then longest wavelength in the paschene series of $L{i}^{+2}$ is:

• A. $\frac{36x}{5}$
• B. $\frac{16x}{7}$
• C. $\frac{9x}{5}$
• D. $\frac{5x}{9}$

Answer 1

The correct option is B $\frac{16x}{7}$

Shortest wavelength in $He+$ (Balmer series)

$\frac{1}{\lambda H_e}=R_H(\frac{1}{2^2}-\frac{1}{{\infty }^2})2^2$

$\frac{1}{\lambda H_e}=\frac{R_H}{4}\times 4$

$\lambda H_e=\frac{1}{R_H}$

$\frac{1}{R_H}=\lambda H_e=x$

Longest wavelength in $L{i}^{+2}$ (paschene series)

$\frac{1}{\lambda H}=R_H{3}^2(\frac{1}{3^2}-\frac{1}{4^2})$

$\frac{1}{\lambda Li}=R_H\left(\frac{7}{144}\right)$

${\lambda }_4=\frac{16}{7R_H}=\frac{16}{7}x$