Question

In certain metal, the stopping potential is found to be 3.70 V. When 235 nm light is incident on the metal, electrons are emitted. What is the maximum kinetic energy given to the electrons in eV and J?

Answer 1

Step-by-step explanation:

d dnnd

Answer 2

3.7 eV

5.92*10^-19 J

Step-by-step explanation:

Given that.

Potential difference of the metal, V = 3.7 V

Wavelength of the light, n = 235 nm

maximum kinetic energy given to the electrons is giving them the formula

K(max) = e.V(s), where

KE(max) is the maximum kinetic energy needed

V = potential difference of the metal

KE(max) = e * 3.7

KE(max) = 3.7eV

converting our answer to Joules, we have

3.7eV = 3.7eV * 1.6*10^-19 J/eV

3.7eV = 5.92*10^-19 J

Therefore, the maximum kinetic energy in both eV and Joules is 3.7eV and 5.92*10^-19 Joules respectively