An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the Photoelectrons. . .

Question : An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the Photoelectrons emitted from this surface have a de-Broglie wavelength λ₁, then prove that, λ = (2mc/h)λ₁²
Doubt by Lakshay & Kulwinder.

Solution :

According to Einstein's Photoelectric Equation

K_max = hν-hν₀
K_max = hν-W₀
W₀ ≈0 (Given)
∴ K_max =hν
K_max = hc/λ
K = hc/λ — (1)

Now, de-Broglie wavelength (λ₁)for particle having Kinetic Energy K
λ₁ = h/(√2mK)
SBS
λ₁² = h²/2mK
Putting the value of K from equation (1)
λ₁²=h²/[2m(hc/λ)]
λ₁² = h²λ /(2mhc)
λ₁² = hλ /(2mc)
2mcλ₁² = hλ
λ = (2mc/h)λ₁²