# Homework Help: The Photoelectric effect, Part 2!

1. Dec 5, 2013

### yaylee

1. The problem statement, all variables and given/known data

Suppose the following: the light source gets very hot (by increasing the intensity), what affect do the heated photons have on photoemission?

Possible choices:
1) they decrease the number of photoelectrons by heating the metal and raising the work function
2) none, since infrared photons are below the cut-off frequency
3) they increase the number of photoelectrons; the hotter the light source, the more effect they have
4) they increase the number of photoelectrons, but by the same amount no matter how hot the light source
5) none, since the infrared photons are absorbed by the air before they reach the metal

2. Relevant equations

The main idea behind this question is understanding that light intensity (increasing or decreasing it) does NOT have an affect on the overall kinetic energy of the photoelectrons. However: by increasing the light intensity, one is essentially increasing the number of INCIDENT light particles on the surface. Hence, the number of electrons released from the metallic surface should be INCREASED.

3. The attempt at a solution

I chose choice 4), because light intensity incidence should be directly related to the number of photoelectrons ejected. This was marked INCORRECT.

Any suggestions or comments would be greatly appreciated!

2. Dec 5, 2013

### physics&math

I believe it would be 3. A more intense light source would mean more photons to liberate more photoelectrons. More photons= more photoelectrons.

Also higher intensity = more photons.

higher intensity = more photons = more photoelectrons, which is what 3 says. 4 says that intensity and the number of photoelectrons are not proportional, which is false.

3. Dec 5, 2013

### physics&math

I believe it would be 3. A more intense light source would mean more photons to liberate more photoelectrons. More photons= more photoelectrons.

Also higher intensity = more photons.

higher intensity = more photons = more photoelectrons, which is what 3 says. 4 says that intensity and the number of photoelectrons are not proportional, which is false.