Question

The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is then

• A. 2.4 eV
• B. 1.5 eV
• C. 3.1 eV
• D. 4.5 eV

Answer 1

The correct option is B 1.5 eV

The minimum negative potential applied to the plate or anode for which the photoelectric current just becomes zero, So, in this case, maximum K.E of an electron will be equal to stopping potential.

Here, $V_0=1.5V,$
Maximum Kinetic energy = $e{V}_0=1.5eV$