Question

Calculate the wavelength of radiation emitted, producing a line in Lyman series, when an electron falls from the fourth stationary state in a hydrogen atom. $(R_H=1.1\times {10}^7{m}^{-1})$

• A. $9.69\times {10}^{-8}metres$
• B. $12.1\times {10}^{-8}metres$
• C. $5169\times {10}^{-8}metres$
• D. $18.1\times {10}^{-8}metres$

Answer 1

The correct option is A $9.69\times {10}^{-8}metres$

$\therefore \frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Given; $R_H=1.1\times {10}^7;$ for Lyman series $n_1=1$ and $n_2=4$(given)
$\therefore \frac{1}{\lambda }=1.1\times {10}^7[\frac{1}{1^2}-\frac{1}{4^2}]$
$\therefore 9.69\times {10}^{-8}metre$