Question

The transition in the hydrogen spectrum, which will have the same wavelength as the Balmer transition, $n=4$ to $n=2$ of $H{e}^{+}$ spectrum, is : (if answer is $n_1=3$ to $n_2=2$, then write as 32)

Answer 1

The transition in the hydrogen spectrum,
which will have the same wavelength as the Balmer transition,
n=4 to n=2 of He+ spectrum, is 12.
(if answer is n1=3 to n2=2, then write as 32)
For Hydrogen sepctrum
$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$......(1)
For He+ spectrum
$\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R_H\times 2^2(\frac{1}{2^2}-\frac{1}{4^2})=4R_H(0.25-0.0625)=0.75R_H$......(2)
From (1) and (2)
$R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})=0.75R_H$
$(\frac{1}{n_1^2}-\frac{1}{n_2^2})=0.75$
Thus, $n_1=1$ and $n_2=2$