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Questions  

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on  C, no photoelectrons will reach the anode A, if the stopping potential of  A relative to C is

a
+3 V
b
+4 V
c
-1 V
d
-3 V

detailed solution

Correct option is D

According Einstein's photoelectric equation kinetic energy of photoelectrons,KEmax=Ev-ϕ or, 2=5-ϕ  ∴ϕ=3eV When Ev=6eV then   KEmax=6-3=3eV or, eVcathode -Vanode =3eV or, Vcathode -Vanode =3 V=-Vstopping ∴Vstopping =-3 V

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