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 λ_d, then :

Concept:

Photosensitive material are materials which eject electrons when photon of frequency greater than their threshold frequency is incident on the surface.

Einstein's photoelectric equation

\(\frac{{hc}}{\lambda }={ϕ _0} + k\)

ϕ0 : work function

k = maximum kinetic energy of photoelectrons

Calculation:

As per Einstein's photoelectric equation

\(\frac{{hc}}{\lambda }={ϕ _0} + k\)

ϕ₀ : work function

k = maximum kinetic energy of photoelectrons

As per question, ϕ → 0

\(\therefore = \frac{{hc}}{\lambda } = k = \frac{{{P^2}}}{{2m}} \Rightarrow P = \sqrt {\frac{{2mhc}}{\lambda }} \)

Now De-broglie wavelength

\({\lambda _d} = \frac{h}{P} = \frac{h}{{\sqrt {2mhc/\lambda } }}\)

\( \Rightarrow \sqrt \lambda = {\lambda _d}\sqrt {\frac{{2mc}}{h}} \)

\( \Rightarrow \lambda = \left( {\frac{{2mc}}{h}} \right)\lambda _d^2\)

Hence Option 4) is correct choice.