Question

In a photoelectric experiment, the potential required to stop the ejection of

electrons from cathode is 4V. What is the value of maximum kinetic energy

of emitted Photoelectrons?​

Answer 1

the maximum kinetic energy is = 4j

Answer 2

Given:

In a photoelectric experiment, the potential required to stop the ejection of electrons from cathode is 4V.

To find:

Maximum kinetic energy of the emitted photoelectrons.

Calculation:

First of all , we need to know a little about PHOTOELECTRIC EFFECT:

• When photons of adequate energy (greater than threshold) falls on a metal surface then electrons are emitted from the surface.

• Now , a certain amount of voltage can be used to stop the ejection of electrons from the metal surface.

• This value of voltage corresponds to the max kinetic energy of the electrons.

So, Max kinetic energy of the electrons:

$\therefore \: KE_{max} = (charge )\times (potential \: difference)$

$= > \: KE_{max} = (e)\times (4)$

$= > \: KE_{max} = 4e \: joule$

So, final answer is:

$\boxed{ \bold{ \large{ \: KE_{max} = 4e \: joule}}}$