Question

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Answer 1

Given: The photoelectric cut-off voltage in a certain experiment is 1.5 V.

The kinetic energy of photoelectrons is given as,

K.E.=e V 0

Where, the stopping potential is V 0 and the charge on the electron is e.

By substituting the values in above equation, we get

K.E.=1.6× 10 −19 ×1.5 =2.4× 10 −19  J

Thus, the maximum kinetic energy of the emitted photoelectron is 2.4× 10 −19  J.