JEE Main On 08 April 2018 Question 9
Question: An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $ {\lambda_n},{\lambda_g} $ be the de Broglie wavelength of the electron in the $ {n^{th}} $ state and the ground state respectively. Let $ {\Lambda_n} $ be the wavelength of the emitted photon in the transition from the $ n^{th} $ state to the ground state. For large n, (A, B are constants) [JEE Main Online 08-04-2018]
Options:
A) $ \Lambda _n^{2}\approx A+B\lambda _n^{2} $
B) $ \Lambda _n^{2}\approx \lambda $
C) $ {\Lambda_n}\approx A+\frac{B}{\lambda _n^{2}} $
D) $ {\Lambda_n}\approx A+B{\lambda_n} $
Show Answer
Answer:
Correct Answer: C
Solution:
De Broglie wavelength $ {\lambda_n} $ $ =\frac{h}{mv}=\frac{h}{\frac{me^{2}}{2n{\in_0}}}=(const)n $
For wavelength of emitted photon $ \frac{hc}{{\Lambda_n}}=13.6( 1-\frac{1}{n^{2}} )eV $
$ {\Lambda_n}=\frac{hc}{13.6}{{( 1-\frac{1}{n^{2}} )}^{-1}}units $
$ =\frac{hc}{13.6}( 1+\frac{1}{n^{2}} )units $ $ =A+\frac{B}{{\lambda_n}^{2}} $ As $ {\lambda_n}\propto n $
