Question

The ratio of wavelengths of photons corresponding to the $\beta$ - line of Lyman series in H - atom and $\alpha$ - line of Balmer series in $L{i}^{2+}$ is:

• A. $45:32$
• B. $9:10$
• C. $32:45$
• D. $3:16$

Answer 1

The correct option is A $45:32$

In Lyman series, $n_1=1$ and beta line corresponds to $n_2=3$
Balmer line for $L{i}^{2+}$, so $Z=3,n_1=2$ and alpha line means $n_2=3$
we know,
$\frac{1}{\lambda }=R_H{Z}^2\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\}$

$\frac{1}{{\lambda }_{Lyman}}=R_H\{1-\frac{1}{9}\}$
${\lambda }_{Lyman}=\frac{9}{8R_H}$

$\frac{1}{{\lambda }_{Balmer}}=R_H\times 9\{\frac{1}{4}-\frac{1}{9}\}$
${\lambda }_{Balmer}=\frac{4}{5R_H}$

$\therefore \frac{{\lambda }_{Lyman}}{{\lambda }_{Balmer}}=\frac{\frac{9}{8R_H}}{\frac{4}{5R_H}}=\frac{45}{32}=45:32$