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?

3.7 eV

5.92*10^-19 J

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

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.