Question

When a hydrogen atom in n=4 state de-excites to n=1 state , the wavelength of the photon emitted $[R=1.097\times {10}^7{m}^{-1}]$

• A. 97.2 nm
• B. 103 nm
• C. 200 nm
• D. 400 nm

Answer 1

The correct option is A 97.2 nm

Wavelength emitted in hydrogen atom $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Given : $n_1=1$ $n_2=4$ $R=1.097\times {10}^7{m}^{-1}$
$\therefore$ $\frac{1}{\lambda }=(1.097\times {10}^7)(\frac{1}{1^2}-\frac{1}{4^2})$
$⟹\lambda =97.2nm$