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?
Praneet Debnath , 8 Years ago
Grade 11
3 Answers
Suraj Prasad
Last Activity: 8 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
Last Activity: 8 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
Last Activity: 5 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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