Question

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n=4$ to $n=2$ of $H{e}^{+}$spectrum

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

Answer 1

Answer: (C)

$H{e}^{+}$ion :
$\frac{1}{\lambda (H)}=R(1{)}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\frac{1}{\lambda \left(H{e}^{+}\right)}=R(2{)}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
Given $\lambda (H)=\lambda \left(H{e}^{+}\right)$
$R(1{)}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]=R(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{1^2}-\frac{1}{2^2}$
On comparing $n_1=1\&n_2=2$