Electrons are emitted from the cathode of a photocell of negligible work function, when photons of wavelength are incident on it. Derive the expression for the de Broglie wavelength of the electrons emitted in terms of the wavelength of the incident light.

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Rupesh Kumar

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E=Φ+K.E max. Where: Φ=work function of electrode Given: Φ=0 ∴ E=K.E of electron where: E= incident energy. λ=h÷p. Where: h= plank cons. P= momentum K.E = (1÷2)mv* *=2 p= mv ∴ p=√2mK.E. Now put value in de- Broglie wavelength formula we get,...
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E=Φ+K.E max. Where: Φ=work function of electrode Given: Φ=0 ∴ E=K.E of electron where: E= incident energy. λ=h÷p. Where: h= plank cons. P= momentum K.E = (1÷2)mv* *=2 p= mv ∴ p=√2mK.E. Now put value in de- Broglie wavelength formula we get, λ = h÷√2mK.E required and. Where:λ= de Broglie wavelength K.E= kinetic energy of electron. read less
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E=Φ+K.E max. Where: Φ=work function of electrode Given: Φ=0 ∴ E=K.E of electron where: E= incident energy. λ=h÷p. Where: h= plank cons. P= momentum K.E = (1÷2)mv* *=2 p= mv ∴ p=√2mK.E. Now put value in de- Broglie wavelength formula we...
read more
E=Φ+K.E max. Where: Φ=work function of electrode Given: Φ=0 ∴ E=K.E of electron where: E= incident energy. λ=h÷p. Where: h= plank cons. P= momentum K.E = (1÷2)mv* *=2 p= mv ∴ p=√2mK.E. Now put value in de- Broglie wavelength formula we get, λ = h÷√2mK.E required and. Where:λ= de Broglie wavelength K.E= kinetic energy of electron. read less
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