Question

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

• A. 1 : 1
• B. 1 : 2
• C. 1 : 4
• D. 3 : 16

Answer 1

The correct option is A 1 : 1

In Lyman series, $n_1=1$ and alpha line means $n_2=2$
Balmer line is for $H{e}^{+}$, so $Z=2,n_1=2$ and beta line means $n_2=4$
we know,
$\frac{1}{\lambda }=R_H{Z}^2\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\}$
$\frac{1}{{\lambda }_{Lyman}}=R_H\{1-\frac{1}{4}\}$
${\lambda }_{Lyman}=\frac{4}{3R_H}$
$\frac{1}{{\lambda }_{Balmer}}=R_H\times 4\{\frac{1}{4}-\frac{1}{16}\}$
${\lambda }_{Balmer}=\frac{4}{3R_H}$
$\frac{{\lambda }_{Lyman}}{{\lambda }_{Balmer}}=\frac{\frac{4}{3R_H}}{\frac{4}{3R_H}}=\frac{1}{1}=1:1$