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 = 4
• B. n = 6 to n = 1
• C. n = 3 to n = 2
• D. n = 2 to n = 1

Answer 1

The correct option is D n = 2 to n = 1

For $H{e}^{+}=2,z=2,\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{2^2}-\frac{1}{4^2}]$
For H, $Z=1\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Since $\lambda$ is same
$4[\frac{1}{2^2}-\frac{1}{4^2}]=[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$n_1=1$ and $n_2=2$