### JEE Main On 16 April 2018 Question 10

##### Question: The de-Broglie wavelength $ ({\lambda_B}) $ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $ ({\lambda_G}) $ by [JEE Main 16-4-2018]

#### Options:

A) $ {\lambda_B}={\lambda _{G/3}} $

B) $ {\lambda_B}={\lambda _{G/2}} $

C) $ {\lambda_B}=2{\lambda_G} $

D) $ {\lambda_B}=3{\lambda_G} $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

- de-Broglie wavelength $ \lambda =\frac{h}{mv} $

$ mvr=\frac{nh}{2\pi } $ $ \frac{h}{mv}=\frac{2\pi r}{n} $

$ \lambda =\frac{2\pi r}{n} $ $ r=a _0\frac{n^{2}}{Z} $

$ \lambda =\frac{2\pi a _0n}{Z} $ As the atom is hydrogen Z = 1 $ {\lambda_B}=\frac{2\pi a _03}{Z} $

$ {\lambda_G}=\frac{2\pi a0}{Z} $

$ {\lambda_B}=3{\lambda_G} $