Question

A spectral line in the spectrum of H atom has a wavenumber of $15222.22c{m}^{-1}$ . The transition responsible for this radiation is (Rydberg constant $R=109677c{m}^{-1}$)

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

Answer 1

The correct option is C

$3\to 2$

Here, let recall what is wave number
$J=\frac{1}{\lambda }\Rightarrow \lambda =\frac{1}{\overline{\nu }}$

And also,
$\overline{\nu }=\frac{1}{\lambda }=R\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Therefore,
$\lambda =\frac{1}{\overline{\nu }}=\frac{1}{15222.22}=6.569\times {10}^{-5}cm=6569\mathring{A}$ (visible light wavelength)
Clearly, it lies in the visible region i.e., in Balmer series. Hence, $n_1=2$. Using the relation for wave number for H atom,

$\overline{\nu }=\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$15222.22=109677(\frac{1}{2^2}-\frac{1}{n_2^2})$

$n_2=3$

The required transition is $3\to 2$.
Hence, (c) is correct.

Note: (d) is wrong because $2\to 3$ will absorb radiation