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 = 2 to n = 1 in H
• B. n = 4 to n = 3 in $H{e}^{+}$
• C. n = 4 to n = 2 in $B{e}^{3+}$
• D. n = 4 to n = 3 in $B{e}^{3+}$

Answer 1

The correct option is A n = 2 to n = 1 in H

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