# In the photoelectric effect ex

In the photoelectric effect experiment, a lightphoton

with a wavelength λ=525 nm hits a metallic cesium

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

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 photoelectronEk

Kinetic energy of the ejected photoelectron Ek =energy of a photon – wave function …………….Equation 1

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

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

= frequency

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

C = speed of light=3 x108 m/s

Thus Energy of photon E = h

= = 3.786x 10-19 J

Energy of photon E= C

Thus

Kinetic energy photoelectron produced Ek

Kinetic energy of the photoelectron produced Ek =energy of a photon – wave function

Ek = E – W

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

Ek =  3.786 x 10-19J

Thus kinetic energy of the photoelectron produced is3.786 x 10-19 J which is equal to theenergy of the photon.

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

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