Question

In the photoelectric effect experiment, a light photon

with a wavelength λ=525 nm hits a metallic cesium

(work function = 3.43 ×10−34 J).\ What is the kinetic energy of the photoelectron produced?

Answer 1

Solution:

A light photon of wavelength = 525 nm hits a metallic Cesium and emits a photoelectron.

Wave function W =  3.43 ×10⁻³⁴ J

wavelength of the light   = 525 nm

To formula to find Kinetic energy of the ejected photoelectron E_k

Kinetic energy of the ejected photoelectron E_k = energy of a photon - wave function ................Equation 1

Lets find Energy of photon E = h,then substitute it in equation 1.

where h = planck's constant = 6.626 x 10^-34 m² Kg/s

= frequency

since frequency of the photon not mentioned we can substitute frequency as ;

C = speed of light=3 x10⁸ m/s

Thus Energy of photon E = h

= = 3.786 x 10^-19 J

Energy of photon E= C

Thus

Kinetic energy photoelectron produced E_k

Kinetic energy of the photoelectron produced E_k = energy of a photon - wave function

E_k = E - W

=  3.786 x 10^-19 J - 3.43 ×10⁻³⁴ J = 3.786 x 10^-19 J

E_k =  3.786 x 10^-19 J

Thus kinetic energy of the photoelectron produced is 3.786 x 10^-19 J which is equal to the energy of the photon.

We can conclude that when a light photon of wavelength = 525 nm hits a metallic Cesium it produces a photoelectron of kinetic energy 3.786 x 10^-19 J .