Question

The wavelength of the radiation emitted when an electron falls from Bohr orbit $4$ to $2$ in hydrogen atom is:

• A. $243nm$
• B. $972nm$
• C. $486nm$
• D. $182nm$

Answer 1

The correct option is C $486nm$

The wavelength of the radiation emitted when an electron falls from Bohr orbit 4 to 2 in a hydrogen atom is 486 nm.

$\frac{1}{\lambda }=1.09677\times {10}^7{m}^{-1}\times [\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=1.09677\times {10}^7{m}^{-1}\times [\frac{1}{2^2}-\frac{1}{4^2}]$

$\frac{1}{\lambda }=1.09677\times {10}^7{m}^{-1}\times 0.1875$

$\frac{1}{\lambda }=2.06\times {10}^6{m}^{-1}$

$\lambda =4.86\times {10}^{-7}m$

$\lambda =486\times {10}^{-9}m$

$\lambda =486nm$
Hence, the option (C) is the correct answer.