If lambdanot is the threshold wavelength for photoelectric emission, lambda is the wavelength of light falling on the surface of metal, and m is the mass of electron, then debroglie wavelength of the emitted electron is
Shahrukh mazumder , 5 Years ago
Grade 12th pass
1 Answers
Arun
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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