Question

The wavelength of the radiation emitted when a hydrogen atom electron falls from infinity to 2nd excited state would be: (Rydberg constant $=1.097\times {10}^7{m}^{-1})$

• A. 406 nm
• B. 192 nm
• C. 821 nm
• D. 168 nm

Answer 1

The correct option is C 821 nm

Second excited state means $n=3$
Using the Rydberg formula,
$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
putting up the values, we get
$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{3^2}-\frac{1}{\infty })=1.218\times {10}^6\lambda =821.02\times {10}^{-9}m\approx 821nm$