Question

Instructions:

Complete the following problems using the format outlined in the homework requirements document.

1. Photons are incident on an Aluminum plate and electrons are emitted as a result of the Photoelectric effect. Provide answers to the following questions:

a. What threshold energy must the photon have to produce a photoelectron from Aluminum, which has a work function of 4.20 eV?

b. Calculate the maximum energy of a photoelectron ejected from Aluminum by UV light with a wavelength of 1500 Angstroms?

c. Does the maximum photoelectron energy vary with the intensity of the UV light (Yes or No)? Explain your answer.

Answer 1

1.Threshold energy is 4.20 eV, because work function is the minimum energy required to remove an electron from the metal surface when a photon incident on metal surface. So either 4.20 eV or greater than 4.20 eV.

2.wavelength =1500 A₀ = 1500 x 10^-10 m.

E= h = h

C= velocity of light = 3 x 10⁸m

h = planks constant = 6.626 x 10^-34

= wavelenght

E= (6.626x10-³⁴) x (3x10⁸) / ( 1500x 10^-10 m)

= 1.3252x10^-18 J

3. The number of electrons ejected is proportional to the intensity or brightness of light. But kinetic energy of the ejected electrons is directly proportional to the frequency of the incident light; these are the basic principles of photoelectric effect.

So the answer is no.

i hope you got the answer