Question

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

• A. $n=4\to n=2$
• B. $n=2\to n=1$
• C. $n=5\to n=2$
• D. $n=3\to n=5$

Answer 1

The correct option is B $n=2\to n=1$

For $H{e}^{+}$ ion, we have
$\frac{1}{\lambda }=Z^2{R}_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=(2{)}^2{R}_H[\frac{1}{(2{)}^2}-\frac{1}{(4{)}^2}]=R_H\frac{3}{4}$
Now for hydrogen atom $\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Equating equations (i) and (ii) we get
$\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}$
Here, $n_1=1$ and $n_2=2$
Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition n = 4 to n = 2 in $H{e}^{+}$ ion