Question

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

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

Answer 1

The correct option is D

n = 2 to n = 1

We know that
Wave number $\overline{\nu }=\frac{1}{\lambda }=[\frac{1}{n_1^2}-\frac{1}{n_2^2}]Z^2\times R$
Therefore,
$\overline{\nu }=\frac{1}{\lambda }=(\frac{1}{2^2}-\frac{1}{4^2})R{Z}^2=\frac{3}{4}R$

In H spectrum for the same $\overline{\nu }=\lambda$ as $Z=1,n=1,n_2=2$