Question

A hydrogen atom emits ultraviolet radition of wavelen gth 102.5 nm. What are the quantum numbers of the states involved in the transition ?

Answer 1

As the light emitted lies in ultraviolet range, the line in lyman series

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

$\Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.1\times {10}^7(\frac{1}{1_2}-\frac{1}{n_2^2})$

$\Rightarrow \frac{{10}^9}{102.5}=1.1\times {10}^7(1-\frac{1}{n_2^2})$

$\Rightarrow \frac{{10}^2}{102.5}=1.1(1-\frac{1}{n_2^2})$

$\Rightarrow 1-\frac{1}{n_2^2}=\frac{100}{102.5\times 1.1}$

$\Rightarrow \frac{1}{n_2^2}=1-\frac{100}{102.5\times 1.1}$

$\Rightarrow n_2=\sqrt{\frac{112.75}{12.75}}=3$

The transition is made 1rom 1 to 3