MCQ
The graph shown below depicts plot of photocurrent versus anode potential for a cathode with 4 eV work function. The energy of the incident photon is:

  • A
    6eV
  • B
    4eV
  • C
    2eV
  • D
    8eV

Answer

  1. 6eV

Explanation:

It is given that

Work function

$\phi$ = 4eV

V = 2V

From Einstein’s relation

hν = $\phi$ + Ek​

Where hν is the energy of photons. And Ek is the kinetic energy

hν = (4+2)eV

hν = 6eV

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