Question
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:

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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