Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be:

[Rydberg constant $=1.097\times {10}^7{m}^{-1}$]

• A. $91nm$
• B. $192nm$
• C. $406nm$
• D. $9.1\times {10}^{-8}nm$

Answer 1

Answer: (A)

$91nm$

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Here $n_1=1,n_2=\infty$

$R=1.097\times {10}^7{m}^{-1}$

$\frac{1}{\lambda }=1.097\times {10}^7[\frac{1}{1^2}-\frac{1}{{\infty }^2}]=1.097\times {10}^7{m}^{-1}\therefore \lambda =9.1\times {10}^{-8}m=91nm$