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}} $ )