Modern Physics 2 Question 12

12. Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ4, the speed of the fastest emitted electron will be

(2016 Main)

(a) >v431/2

(b) <v431/2

(c) =v431/2

(d) =v341/2

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Solution:

  1. According to the law of conservation of energy, i.e. Energy of a photon (hν)= Work function (φ)+ Kinetic energy of the photoelectron 12mvmax2

According to Einstein’s photoelectric emission of light i.e.

E=(KE)max +φhcλ=(KE)max+φ

As,

If the wavelength of radiation is changed to 3λ4, then

43hcλ=43(KE)max+φ3+φ

(KE)max for fastest emitted electron =12mv2+φ

12mv2=4312mv2+φ3 i.e. v>v43



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