Question 31:
easy
An electromagnetic wave of wavelength \(lambda\) is incident on a photosensitive surface of negligible work function. If \(m\) is mass of photoelectron emitted from the surface has de-Broglie wavelength \(lambda_d\), then
With a negligible work function, the maximum kinetic energy of the emitted photoelectron is \(E = \frac{hc}{\lambda}\) and its de-Broglie wavelength is \(\lambda_d = \frac{h}{\sqrt{2mE}}\). Substituting \(E\) gives \(\lambda_d^2 = \frac{h\lambda}{2mc}\), which simplifies to \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\).