Is My Photoelectric Effect Calculation Correct?

[kraphysics]

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?

 

[ehild]

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

 

[kraphysics]

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

 

[ehild]

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

 

[kraphysics]

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?

 

[SammyS]

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: E_p = h·f , where h is Planck's constant.

 

[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?

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

 

[kraphysics]

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.