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 April 16, 2024 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.

Answer.

Transcript

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.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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