Am electromagnetic wave of wavelength lambda is incident on a photosensitive surface of negligible work function If the photoelectrons - Physics - Dual Nature Of Radiation And Matter

Question

Am electromagnetic wave of wavelength lambda is incident on a
photosensitive surface of negligible work function If the
photoelectrons emitted from this surface have de-broglie wavelength
lambda 1, prove that lambda= (2mc/h)*lambda1^2

Vara · Accepted Answer

Dear
Student,
Please find below the
solution to the asked query:
Debroglie wavelength:λB=h2mEλB2=h22mEwhereE=Energy=hcλSo,λB2=h22mhcλλ=2mchλB2
Hope this information will
clear your doubts about Debroglie
wavelength
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Am electromagnetic wave of wavelength lambda is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have de-broglie wavelength lambda 1, prove that lambda= (2mc/h)*lambda1^2