Question

A photon of energy 8 eV is incident on a metal
surface of threshold frequency 1.6 x 1015 Hz.
The kinetic energy of the photoelectrons emitted
in eV is ſh = 6.63 x 10-34
Js]​

Answer 1

Answer:

0.192 × 10^(-19) joulee

Explanation:

According there is a formula

$hv \: - hv_{0} = \frac{1}{2 } \: {m(v)}^{2}$

Also 1 ev = 1.6 × 10^(-19) joules

Thus hv = 8 * ev

Also other values are given