Question

The threshold wavelength of a given photosensitive surface is λ₀. When light of wavelength λ(= λ₀/2) is incident on this photosensitive surface, the ‘stopping potential’, needed to ‘stop’ the most energetic emitted photoelectrons, is V.

If λ_B denotes the deBroglie wavelength, associated with an electron, accelerated through a potential V, the ratio (λ_B/√λ₀) equals

Answer 1

Correct option :  (3)

Explanation :

Energy of a photon of the incident light = hc/λ

Work function = Energy of a photon of light, of wavelength equal to the threshold wavelength(λ₀) = hc/λ₀ 

∴ If V be the stopping potential, we have

The deBroglie wavelength, associated with an electron, accelerated through a potential