Did I solve this photoelectric problem right?

In summary, the photoelectric effect is an interaction between an electron and a photon, where the electron gains the energy of the photon in order to leave the metal surface. The work function is the minimum energy required for this process to occur. The power or intensity of the light source does not affect the maximum kinetic energy of the emitted electrons, as it depends only on the energy of the photons. However, for a monochromatic light source, the number of electrons emitted may be directly proportional to the intensity of the light.
  • #1
kraphysics
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0

Homework Statement


Light of wavelength 5.4 * 10^-7 m falls on a photosurface and causes the emission of electrons of maximum kinetic energy 2.1 eV at a rate of 10^15 per second. The light is emitted by a 60 W light bulb.



Homework Equations


Find the work function of the surface.


The Attempt at a Solution


Ok I actually have an answer. I'm not sure if it's right though. I just want you guys to check if my reasoning and answer makes sense.
So first I thught since P= W/t, 60W = 60 J /s
That means more than one photon is hitting the surface. Then, E = W + Ek
60 J - ((2.1eV)(1.60*10^-19)(10^15 electrons)) = W
59.999664 J,
Then I thought that the work function means the amount of energy required to remove just one electron so, I divided 59.999664 J by (1 * 10^15 electrons) and got 5.9999664 * 10^-14 J.
Does this make sense?
 
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  • #2
No, the photoelectric effect is an interaction between an electron and a photon. The electron gets the energy of the photon, hf, and if it is greater then the work function W it can leave the metal. The difference of the photon energy and work function is the kinetic energy of the electron. KE=hf-W. Electrons at quite on the surface leave with this maximum kinetic energy, those a bit under the surface can loose some of their energy, that is why the "maximum kinetic energy" is mentioned. If electrons are emitted or not from a given material depends only on the energy of the photon, not the intensity of light. ehild
 
  • #3
So the power of the light bulb doesn't matter? So then would it just be:

(6.63 * 10^-34 J*s) ( (3*10^8)/(5.4 *10^-7m)) - (2.1 eV)(1.60 * 10^-19) = W
W = 3.68 * 10^-19 J
 
  • #4
The power of the light bulb does not matter. Think: The light bulb emits different light with different wavelengths, red, yellow, blue, even infrared (heat) only a little part of its radiation is around the desired wavelength. But even in the case of monochromatic light source, the intensity of light which is proportional to the number of photons hitting the metal, will only increase the number of photons emitted in unit time.

ehild
 
  • #5
ehild said:
The power of the light bulb does not matter. Think: The light bulb emits different light with different wavelengths, red, yellow, blue, even infrared (heat) only a little part of its radiation is around the desired wavelength. But even in the case of monochromatic light source, the intensity of light which is proportional to the number of photons hitting the metal, will only increase the number of photons emitted in unit time.

ehild

Thanks. Just an additional questional that I have. Is it safe to say that the intensity(power) of the light source is directly proportional to the number of number of electrons emitted? So say, if the intensity of the light source was 120 W instead of 60 W, then the number of electrons released would double as well?
 
  • #6
Hello kraphysics.

It appears to me that your solution assumes that ALL of the energy converted by the light bulb goes into freeing electrons from the surface. -- that's highly unlikely.

The energy of each photon needs to be greater than the work function to free an electron. Any excess energy goes to the electron's Kinetic Energy.

The energy of a photon is proportional to it's frequency: Ep = h·f , where h is Planck's constant.
 
  • #7
kraphysics said:
Thanks. Just an additional questional that I have. Is it safe to say that the intensity(power) of the light source is directly proportional to the number of number of electrons emitted? So say, if the intensity of the light source was 120 W instead of 60 W, then the number of electrons released would double as well?
If you have a monochromatic light source you can say that the number of electrons emitted is about proportional to the intensity of light. See: http://en.wikipedia.org/wiki/Photoelectric_effect. But the intensity of the light of appropriate wavelength is not directly proportional to the power of the light bulb, it depends on other factors, for example, the temperature of the tungsten wire in the bulb.

ehild
 
Last edited:
  • #8
Thanks a lot ehild. Yea for my problem, we don't have to worry about temperature. I just wanted to know if the number of photoelectrons emitted is directly proportional to number of photons(intensity of light). It seems like that's true.
 

1. How do I know if I solved a photoelectric problem correctly?

The best way to know if you solved a photoelectric problem correctly is to check your answer against the correct solution. This can be done by consulting a textbook or asking a teacher or fellow scientist for confirmation. Additionally, you can double-check your calculations and make sure they are accurate.

2. What are some common mistakes to avoid when solving photoelectric problems?

Some common mistakes to avoid when solving photoelectric problems include forgetting to convert units, using incorrect values for constants, and making errors in mathematical calculations. It is important to always carefully check your work and make sure all steps are correct.

3. Can I use the same approach for solving all photoelectric problems?

No, each photoelectric problem may require a slightly different approach depending on the specific variables and conditions given. It is important to carefully read and understand the problem before attempting to solve it.

4. Is there a formula for solving photoelectric problems?

Yes, there is a formula for solving photoelectric problems. The most commonly used formula is the photoelectric effect equation, which is given by E = hf - φ, where E is the energy of the incident photon, h is Planck's constant, f is the frequency of the incident light, and φ is the work function of the material.

5. Are there any resources available to help me solve photoelectric problems?

Yes, there are many resources available to help you solve photoelectric problems. These include textbooks, online tutorials, and practice problems with solutions. Additionally, you can consult with a teacher or fellow scientist for guidance and clarification on any questions you may have.

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