The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Photoelectric cut-off voltage,Vo=1.5V The maximum kinetic energy of the emitted photoelectrons is given as- Ke=eVo Ke=1.6×10−19×1.5 J ⇒2.4×10−19 J Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is2.4×10−19 J
