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If λ 0 is the threshold wavelength for photoelectric emission, λ wavelength of light falling on the surface of metal, and m, mass of electron, the de Broglie wavelength of emitted electron is?

If λ0 is the threshold wavelength for photoelectric emission, λ wavelength of light falling on the surface of metal, and m, mass of electron, the de Broglie wavelength of emitted electron is?

Grade:11

3 Answers

Suraj Prasad IIT Patna
askIITians Faculty 286 Points
7 years ago
see , hλ = hλ0 + 1/2(mv^2)

K.E = h(λ –λ0)

de broglie’s wavelength attached to itλ = h/p =h/sqrt(2 K.E*m)
Hence ,λ = h/sqrt(2*h(λ-λ0)*m)
I hope you got it.
Praneet Debnath
33 Points
7 years ago
The Correct Answer is { [hλλ0] / [2mc(λ0-λ)] }. 
BUt the question is how is it coming so...??
I want a detailed answer please.
Somebody please kindly refer to my request.
 
 
Tanisha Singh
20 Points
4 years ago
 hc/λ = hc/λ0 + 1/2(mv^2)
 
K.E = hc(1 /λ –1 /λ0)
 
de broglie’s wavelength attached to itλ = h/p =h/sqrt(2 K.E*m)
So, 
λ = h/(2hc(λ0 - λ) m/λ0λ)½
 
λ  = (h(λ0λ) /2mc(λ0 - λ))1/2
 
 

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