Question

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

• A. $n_1=3,n_2=6$
• B. $n_1=2,n_2=4$
• C. $n_1=1,n_2=2$
• D. $n_1=1,n_2=4$

Answer 1

The correct option is C $n_1=1,n_2=2$

For $He+$
$\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{2^2}-\frac{1}{4^2}]=R_H{2}^2[\frac{1}{2^2}-\frac{1}{4^2}]$
Let transition in $H$-atom across from $n_2$ to $n_1$
$\therefore$ for $H;\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ $=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Since, $\lambda$ is same for both the transitions
$\therefore 2^2[\frac{1}{2^2}-\frac{1}{4^2}]=[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
or $\frac{1}{1^2}-\frac{1}{2^2}=[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\therefore n_1=1,n_2=2$