Question

Calculate the wavelength of the radiation emitted producing a line in the 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.
  $\lambda =650.5\stackrel{o}{A}$
• B. $\lambda =919.7\stackrel{o}{A}$
  
• C. $\lambda =969.6\stackrel{o}{A}$
  
• D. $\lambda =848.2\stackrel{o}{A}$

Answer 1

The correct option is C $\lambda =969.6\stackrel{o}{A}$


We know,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
where $R_H$ is the Rydberg constant
$\lambda$ = wavelength of the radiation
$\frac{1}{\lambda }=1.1\times {10}^7\times 1^2[\frac{1}{1^2}-\frac{1}{4^2}]\therefore \lambda =0.9696\times {10}^{-7}$
$\lambda =969.6\stackrel{o}{A}$