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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. 

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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.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.