Question

Obtain Bohr’s quantisation condition for angular momentum of electron orbiting in nth orbit in hydrogen  atom on the basis of the wave picture of an electron using de Broglie hypothesis. 

Answer 1

`n lambda = 2 pi r`

But ` lambda = n/(mv) `( By De Broglie)

`(n lambda) /(mv) = 2 pi r`

` therefore (lambda n) /(2 pi ) = mvr `        (mvr =  L)

` therefore L = (lambda n)/(2 pi)`

Answer 2

Bohr's second postulate states that the angular momentum of an electron has only those values that are integral multiples of `"h"/(2pi)` He thought that the motion of electrons within an atom is associated with the standing wave along the orbit as shown.

About standing waves in stretched strings, we know that only those waves survive for which the distances traveled in the round trip between the ends are integral multiples of the wavelength. Similarly, for an electron moving in the nth orbit of radius r_n, the distance traveled in one trip is 2πr_n, which should be an integral multiple of the wavelength.

2πr_n = nλ (where n= 1, 2, 3, 4 etc)

By de-Broglie hypothesis, we have:

`lambda = "h"/"p" = "h"/("mv"_"n")`

Substituting the value of λ in the above expression, we get:

`2pi"r"_"n" = "n" "h"/"mv"_"n"`

⇒ `"mv"_"n""r"_"n" = "n" "h"/(2pi)`

Angular momentum `= "L" = "n" "h"/(2pi)`