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Question
According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?
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Solution
Let v0 be the velocity of the electron moving in the ground state and r0 be the radius of the ground state.
Frequency of the revolution of electron in the circle is given by
`f = V_0/(2pir_0)`
Frequency of the radiation emitted = Frequency of the revolution of electron
`therefore ` Frequency of the radiation emitted = `V_0/(2pir_0)`
Also, c = `flamda`
Here, c = Speed of light
`lamda` = Wavelength of the radiation emitted
⇒ `lamda = c/f`
`therefore lamda = (2pir_0c)/V_0`
`= (2xx(3.14)xx(53xx10^-12)xx(3xx10^8))/(2.187xx 10^6)`
= 45.686 × 10-12m = 45.7 nm
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