Gujarati
Hindi
12.Atoms
normal

An electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of tha hydrogen atom. The wave number of the emitted radiations ($R =$ Rydberg's constant) will be

A

$\frac{16}{3R}$

B

$\frac{2R}{16}$

C

$\frac{3R}{16}$

D

$\frac{4R}{16}$

Solution

Wave number

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 \mathrm{R}}{16}$

Standard 12
Physics

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