Question

Wavelength of a light emitted from second orbit to first orbit in a hydrogen atom is:

• A. $1.215\times {10}^{-7}m$
• B. $1.215\times {10}^{-5}m$
• C. $1.215\times {10}^{-4}m$
• D. $1.215\times {10}^{-3}m$

Answer 1

Answer: (A)

Here: initial orbit $n_1=1$ Final orbit $n_2=2$ The relation for the wavelength is $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left[\frac{1}{{(1)}^2}-\frac{1}{{(2)}^2}\right]=\frac{3R}{4}$ $\lambda =\frac{4}{3R}=\frac{4}{3(1.097\times {10}^7)}$ $=1.215\times {10}^{-7}m$ (where Rydberg constant $R=1.097\times {10}^7{m}^{-1}$ )