Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, $n=4$ to $n=2$ of ${He}^{+}$ spectrum?

• A. $n=2$ to $n=1$
• B. $n=4$ to $n=2$
• C. $n=3$ to $n=2$
• D. $n=4$ to $n=3$

Answer 1

Answer: (A)

We have to compare wavelength of transition in the $H$-spectrum with the Balmer transition $n=4$ to $n=2$ of ${He}^{+}$ spectrum.
$\because {\lambda }_H={\lambda }_{{He}^{+}}$
$\therefore R_H{Z}_H^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=R_H{Z}_{{He}^{+}}^2[\frac{1}{2^2}-\frac{1}{4^2}]$
$1\times [\frac{1}{n_1^2}-\frac{1}{n_2^2}]=4\times (\frac{1}{4}-\frac{1}{16})$
$[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=4\times \frac{4-1}{16}$
$[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=\frac{3}{4}$
If $n_1=1$, then $n_2=2,3,...$
For first line $n_2=2$, $n_1=1$
$[\frac{1}{1^2}-\frac{1}{2^2}]=\frac{1}{1}-\frac{1}{4}=\frac{3}{4}$
Hence, transition $n_2=2$ to $n_1=1$ will give spectrum of the same wavelength as that of Balmer transition, $n_2=4$ to $n_1=2$ in ${He}^{+}$.