In the Bohr model of hydrogen atom. What is the de-Broglie wavelength `lambda` for the electron when it is in the (i) n = 1 level (ll) n = 4 level. In each case, compare the de-Broglie wave length to the circumference of the orbit.
Data: n=1, n=4,`lambda`=?

Explanation

By Bohr.s first postulate,
angular momentum of the electron `=(nh)/(2pi)`
`mvr=(nh)/(2pi)`
`:. (2pir)/(n)=(h)/(mv)` ...(1)
But, (de brogile wave length),
`lambda=(h)/(mv)` ...(2)
From the equation (1) and (2)
`lambda=(2pir)/(n)` ...(3)
(i) `n=1, r=r_(1)=0.53 Å`
`:. lambda_(1)=2pir_(1)=2pi(0.53)Å=3.328Å`
From equation(3), `(lambda)/(2pir)=(1)/(n),` for n=1,
`(lambda)/(2pir)=1`, i.e., `lambda-2pir`
`:.` In first orbit, de-brogile wave length of the electron is equal to the circumference of the orbit.
(ii) n = 4, `r_(4)=4^(2)r_(1)`
from equation (3), `lambda_(4)=(2pir_(4))/(4)=(2pi(4^(2)r_(1)))/(4)`
i.e., `lambda_(4)=2pixx4xx(0.53)Å`
`lambda_(4)=13.313Å`
and from `lambda_(4)=(2pir_(4))/(4)`, the de-Brogile wavelength of the electron in the fourth orbit is equal to one fourth of the perimeter.