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

The correct answer is: $2.4×10^{−19}J.$
Photoelectric cut-off voltage,$V_0=1.5 V$
The maximum kinetic energy of the emitted photoelectrons is given as:
$K_e=eV_0$
Where,
e=Charge on an electron$=1.6×10^{−19}C$
$∴K_e=1.6×10^{−19}×1.5$
$=2.4×10^{-19}J$
Therefore,the maximum kinetic energy of the photoelectrons emitted in the given experiment is $2.4×10^{−19}J.$