Question

The de-Broglie wavelength $ \lambda $ of an electron revolving in $ {4^{th}} $ orbit of radius $ r $ in hydrogen atom is given by
(A) $ \lambda = \dfrac{{\pi r}}{4} $
(B) $ \lambda = \dfrac{{\pi r}}{2} $
(C) $ \lambda = \dfrac{{3\pi r}}{2} $
(D) $ \lambda = \dfrac{{3\pi r}}{4} $

Answer 1

Hint : To solve this question, we have to use the second postulate of the Bohr’s model of atom from which the momentum of the electron in the orbit would be determined. Then using de-Broglie's relation we can get the final answer.

Formula used: The formulae which have been used in this solution are given by
 $ L = \dfrac{{nh}}{{2\pi }} $ , here $ L $ is the angular momentum of an electron revolving in the $ {n^{th}} $ orbit of an atom.
 $ L = mvr $ , here $ L $ is the angular momentum of a body of mass $ m $ moving with a speed of $ v $ in a circle of radius $ r $ .
 $ \lambda = \dfrac{h}{p} $ , here $ \lambda $ is the wavelength of a body having a momentum of $ p $ , and $ h $ is the Planck’s constant.

Complete step by step answer
We know that the de-Broglie wavelength of a particle is given by
 $ \lambda = \dfrac{h}{p} $ (1)
So we need the momentum of the electron which is revolving in the $ {4^{th}} $ orbit of the hydrogen atom. For this we use the second postulate of Bohr’s model which defines the angular momentum of the electron as
 $ L = \dfrac{{nh}}{{2\pi }} $
We know that $ L = mvr $ . So we have
 $ mvr = \dfrac{{nh}}{{2\pi }} $
Dividing by the radius $ r $ on both the sides, we get
 $ mv = \dfrac{{nh}}{{2\pi r}} $
We know that the momentum $ p = mv $ . So we have
 $ p = \dfrac{{nh}}{{2\pi r}} $ (2)
Substituting (2) in (1) we have
 $ \lambda = \dfrac{{2\pi rh}}{{nh}} $
On simplifying we get
 $ \lambda = \dfrac{{2\pi r}}{n} $
Since the electron is revolving in the $ {4^{th}} $ , so we have $ n = 4 $ . Therefore we finally get
 $ \lambda = \dfrac{{2\pi r}}{4} $
 $ \Rightarrow \lambda = \dfrac{{\pi r}}{2} $
Thus, the de-Broglie wavelength $ \lambda $ of an electron revolving in $ {4^{th}} $ orbit of radius $ r $ in hydrogen atom is equal to $ \dfrac{{\pi r}}{2} $ .
Hence the correct answer is option (B).

Note
We should not try to obtain the wavelength in the exact numeric form by computing the radius of the electron in the fourth orbit of the hydrogen atom. Although we can still get the correct answer, doing so will only make our calculations lengthy.