Photoelectric Surface: Work Function & Kinetic Energy

[potamus]

Homework Statement

A certain photoelectric surface has a threshold frequency of 4.62 X 10^14 Hz.
a) What is the photoelectric work function in joules?
b) The surface is illuminated with light that has a frequency of 5.2 X 10^14 Hz. What is the kinetic energy of the electrons ejected from the surface? Answer in joules and electron volts.

Homework Equations

1 eV = 1.6 X 10^19 J
W = hf

The Attempt at a Solution

a) 3.06 X 10^-19 J
b) I need help on! I have no idea which equation to use, etc. Thanks!

 

[Hootenanny]

Homework Statement

A certain photoelectric surface has a threshold frequency of 4.62 X 10^14 Hz.
a) What is the photoelectric work function in joules?
b) The surface is illuminated with light that has a frequency of 5.2 X 10^14 Hz. What is the kinetic energy of the electrons ejected from the surface? Answer in joules and electron volts.

Homework Equations

1 eV = 1.6 X 10^19 J
W = hf

The Attempt at a Solution

a) 3.06 X 10^-19 J
b) I need help on! I have no idea which equation to use, etc. Thanks!

Welcome to Physics Forums,

You're answer to part (a) is spot on. Now for part (b), what is the definition of the work function?

 

[thomate1]

I would like to first clarify the concept of photoelectric effect and how it relates to the given problem. The photoelectric effect is the phenomenon where electrons are emitted from a metal surface when it is exposed to light of sufficient frequency. The frequency of the light is directly proportional to the energy of the photons, and if the frequency is above a certain threshold value, the electrons will be ejected from the surface with a certain kinetic energy.

Now, let's address the questions.

a) The photoelectric work function (W) is the minimum amount of energy required to remove an electron from the surface. It is given by the equation W = hf, where h is the Planck's constant and f is the frequency of the incident light. Since the threshold frequency is given as 4.62 X 10^14 Hz, we can calculate the work function as follows:

W = (6.626 X 10^-34 J s)(4.62 X 10^14 Hz) = 3.06 X 10^-19 J

b) To calculate the kinetic energy of the ejected electrons, we can use the equation K.E. = hf - W, where K.E. is the kinetic energy, h is the Planck's constant, f is the frequency of the incident light, and W is the work function. Since the frequency of the incident light is given as 5.2 X 10^14 Hz, we can calculate the kinetic energy as follows:

K.E. = (6.626 X 10^-34 J s)(5.2 X 10^14 Hz) - 3.06 X 10^-19 J = 3.42 X 10^-19 J

To convert this to electron volts, we can use the conversion factor 1 eV = 1.6 X 10^-19 J. Therefore, the kinetic energy in electron volts would be:

K.E. = (3.42 X 10^-19 J)/(1.6 X 10^-19 J/eV) = 2.14 eV

In conclusion, the kinetic energy of the ejected electrons from the given photoelectric surface when illuminated with light of frequency 5.2 X 10^14 Hz is 3.42 X 10^-19 J or 2.14 eV.