The photoelectric cut-off voltage in an experiment was found to be 1.5 V. The work function for the material used in the experiment was 4.2 eV. The maximum kinetic energy of the photoelectrons that was emitted was:

Question

The photoelectric cut-off voltage in an experiment was found to be 1.5 V. The work function for the material used in the experiment was 4.2 eV. The maximum kinetic energy of the photoelectrons that was emitted was: (A) 1.5 eV (B) 2.7 eV (C) 4.2 eV (D) 5.7 eV

Tardigrade Admin · Accepted Answer

If stopping potential is V0, then maximum kinetic energy of photoelectrons is given by Ek=eV Given, V= 1.5 volt ∴ Ek=1.5eV

Q. The photoelectric cut-off voltage in an experiment was found to be 1.5 V. The work function for the material used in the experiment was 4.2 eV. The maximum kinetic energy of the photoelectrons that was emitted was:

1.5 eV

2.7 eV

4.2 eV

5.7 eV

none of the above

If stopping potential is $V_{0},$ then maximum kinetic energy of photoelectrons is given by $E k = e V$ Given, V= 1.5 volt $∴$ $E_{k} = 1.5 e V$