Question

If the shortest wavelength of the spectral line of H - atom in Lyman series is x, then the longest wavelength of the line in Balmer series of $L{i}^{2+}$ is

• A. $\frac{5x}{4}$
• B. $\frac{4x}{5}$
• C. $\frac{x}{9}$
• D. $9x$

Answer 1

The correct option is B

$\frac{4x}{5}$

$v_{max}=\frac{1}{{\lambda }_{min}}=R_H\frac{Z^2}{n_1^2}(n_2=\infty )$
For H - atom, Z = 1 : For Lyman series, $n_1=1$
Hence, $\frac{1}{x}=R_H$
For largest wavelength in the Balmerseries of $L{i}^{2+},Z=3,n_1=2;n_2=3$
Hence, $\frac{1}{{\lambda }_{ma{x}^{\prime }}}{R}_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$=\frac{1}{x}\times 3^2(\frac{1}{2^2}-\frac{1}{3^2})=\frac{9}{x}\left[\frac{5}{36}\right]=\frac{5}{4x}$
or ${\lambda }_{max}=\frac{4x}{5}$