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

Dear Student,
Please find below the solution to the asked query:
 Debroglie wavelength:λB=h2mEλB2=h22mEwhereE=Energy=hcλSo,λB2=h22mhcλλ=2mchλB2
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