Question

Which jump is responsible for the wave number of emitted radiations equal to $9.7490\times {10}^6{m}^{-1}$ in Lyman series of hydrogen spectrum? $(R=1.09678\times {10}^7{m}^{-1})$

Answer 1

Wave number $=\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{n_2^2}]$ for Lyman series.

$\therefore 9.7490\times {10}^6=1.09678\times {10}^7[1-\frac{1}{n_2^2}]0.888=1-\frac{1}{n_2^2}\therefore n_2\approx 3$

Jump from $n_1=1$ to $n_2=3$ is responsible for given wave no.