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13/09/2021 11:35 am
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An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then:
(1) λ = \((\frac{2m}{hc})\lambda_d^2\)
(2) λd = \((\frac{2mc}{h})\lambda^2\)
(3) λ = \((\frac{2mc}{h})\lambda_d^2\)
(4) λ = \((\frac{2h}{mc})\lambda_d^2\)
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13/09/2021 11:42 am
Correct answer: (3) λ = \((\frac{2mc}{h})\lambda_d^2\)
Explanation:
\(\frac{hc}{\lambda}\) = Kmax + ϕ[given ϕ is neglibible]
so, \(\frac{hc}{\lambda}\) = Kmax
λd = \(\frac{h}{\sqrt{2m\;K_{max}}}\)
⇒ Kmax = \(\frac{h^2}{2m \lambda_d^2}\)
\(\frac{hc}{\lambda}\) = \(\frac{h^2}{2m \lambda_d^2}\)
⇒ λ = \((\frac{2mc}{h})\lambda_d^2\)