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?
Answer:
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 .
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
![](https://writinghelpe.com/wp-content/uploads/2022/08/save.jpg)