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

Photoelectric cut-off voltage,$V_o=1.5V$

The maximum kinetic energy of the emitted photoelectrons is given as-

$K_e=e{V}_o$

$K_e=1.6\times {10}^{-19}\times 1.5J$

$\Rightarrow 2.4\times {10}^{-19}J$

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is$2.4\times {10}^{-19}J$