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ADNAN MUHAMMED

Grade 12,

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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

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

Grade:12th pass

1 Answers

Arun
25750 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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