CBSE Class 12 Sample Paper for 2022 Boards [Term 2] - Physics

Question 4 - CBSE Class 12 Sample Paper for 2022 Boards [Term 2] - Physics - Solutions to CBSE Sample Paper - Physics Class 12

Last updated at Jan. 28, 2022 by Teachoo

Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals the classical frequency of revolution of an electron.

Question 4
Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals the classical frequency of revolution of an electron.
The hydrogen atom is de-exciting from an upper level (n) to a lower level (n − 1).
The spatial frequency of emitted radiation as electron de-excited from level n2 to n1 is given by:
𝑓=(2𝜋^2 𝑚𝑘^2 𝑧^2 𝑒^4)/ℎ^3 [1/𝑛^2 −1/𝑛^2 ]
This is also known as the wave number.
When the electron travels from n to n-1:
For hydrogen atom z = 1
𝑓=(2𝜋^2 𝑚𝑘^2 𝑒^4)/ℎ^3 [1/〖(𝑛 − 1)〗^2 −1/𝑛^2 ]
=(2𝜋^2 𝑚𝑘^2 𝑒^4)/ℎ^3 [(𝑛^2 − 𝑛^2 −1 + 2𝑛)/(𝑛^2 (𝑛 − 1)^2 )]
=(2𝜋^2 𝑚 𝑘^2 𝑒^4)/ℎ^3 [(2𝑛 − 1)/(𝑛^2 (𝑛 − 1)^2 )]
For large value of −n,
2n − 1 ~ 2n
n − 1 ~ n
𝑓=(2𝜋^2 𝑚𝑘^2 𝑒^4 (2𝑛))/(ℎ^3 (𝑛^4))=(4𝜋^2 𝑚 𝑘^2 𝑒^4)/〖𝑛^3 ℎ〗^3
The orbital frequency of revolution of an electron in
Nth orbit is given by:
𝑓=𝑉/2𝜋𝑟= (4𝜋^2 𝑚𝑘^2 𝑒^4)/(𝑛^3 ℎ^3 )
Both values are equal. Hence proved.

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.