The photoelectric cut off voltage in a certain experiment is 1.5V. What is the maximum kinetic energy of photoelectrons emitted? `e=1.6xx10^(-19)C`.

Explanation

Photoelectric cut-off voltage , `V_0=1.5 V`
The maximum kinetic energy of the emitted photoelectrons is given as : `K_c=eV_0`
Where, e=Charge on an electron `=1.6xx10^(-19)C`
`therefore K_c=1.6xx10^(-19) xx1.5`
`=2.4xx10^(-19) J`
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is `2.4xx10^(-19) J`