The Photoelectric effect, Part 2

AI Thread Summary
Increasing the intensity of a light source leads to a higher number of incident photons, which in turn increases the number of photoelectrons emitted from a metallic surface. The discussion clarifies that while light intensity does not affect the kinetic energy of the photoelectrons, it directly correlates with the number of photons available to liberate electrons. The consensus leans towards choice 3, which states that a hotter light source results in more photoelectrons due to the increased number of photons. Choice 4 is deemed incorrect as it suggests that intensity and photoelectron emission are not proportional. Overall, the relationship between light intensity and photoemission is emphasized as a key concept.
yaylee
Messages
22
Reaction score
0

Homework Statement



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

Homework 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.

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!
 
Physics news on Phys.org
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.
 
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top