# Photoelectric effect

1. Sep 10, 2010

### fluidistic

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

A monochromatic light beam with wavelength $$\lambda =405 nm$$ and intensity $$3 \times 10 ^{-9}\frac{W}{m^2}$$ points to one of the plates of a plane metallic capacitor (photocathode). We apply a difference of potential of 1.15V between the 2 plates in such a way that we slow down the photoelectrons. A pico-ammeter shows a null current.
1)What's the maximum kinetic energy of the electrons?
2)What's the work function of the metal?
3)How many photoelectrons are emited by unit of area over unit of time?
4)What's the maximum wavelength of the radiation that can detach electrons from the metal?

2. The attempt at a solution
Not sure, but I believe that the maximum kinetic energy is $$E_K=hf-\phi$$ where $$\phi$$ is the work function and is worth $$hf_0$$. That is, according to http://en.wikipedia.org/wiki/Photoelectric_effect.

What confuses me is the difference of potential of 1.15V.
Anyway, if the pico-ammeter shows $$0 \times 10 ^{-12}A$$, it means that the smallest current it can measure is $$1 \times 10 ^{-12}A$$. I'm sure I need this data in order to solve part 1). I also think I'd solve part 2) before part 1)...
I really don't know how to solve the problem.

2. Sep 10, 2010

### rock.freak667

The pd is doing work, such that it is reducing the speed of the electrons from maximum to zero.

Work done by pd = change in kinetic energy.

3. Sep 10, 2010

### fluidistic

Thanks for helping!
Ah ok! Good to know... I didn't know that.
Ok so I must find the work done by the battery or generator, etc. I think it is 1.15V divided by the total charge (in coulomb). So should I find the number of electrons detached from the metal?
This would mean that the battery does more work when there are more photoelectrons, despite the voltage stays constant. Is this correct?

4. Sep 10, 2010

### rock.freak667

no no, to find the work done by the battery on the electrons, you just need to multiply the charge of one electron by the pd. W=VQ.

5. Sep 10, 2010

### fluidistic

Ah ok. Then it gives me $$1.8425 \times 10 ^{-19}J$$. So this is the answer to part a)? Namely the maximum kinetic energy of any electron (a single one, as opposed to all the electrons).

6. Sep 10, 2010

### rock.freak667

Yes that would be correct, but that gives the maximum KE of all the electrons.

7. Sep 11, 2010

### fluidistic

Do you mean that it gives the sum of the maximum kinetic energy of each electrons? If so, then I don't really understand why.

8. Sep 11, 2010

### rock.freak667

The pd is working to slow down all of the electrons, not just one.

9. Sep 11, 2010

### fluidistic

Ok I get you.
I find hard to imagine the situation. When the photons hit the metal and electrons, where do the electrons go? Do they stick within the circuit or are they ejected between the plates of the capacitor?
What if I modify the generator to be 10V? Would the electrons slow down and then go fast (with an acceleration) backward compared to their motion when they get detached, so that they hit the metal of the capacitor? That is, in the case the electrons get ejected between the plates of the capacitor.

10. Sep 11, 2010

### rock.freak667

I am not too sure as to 'where' they go, but they are emitted from the surface of the metal, so connecting an ammeter will record the current. If you modify the voltage, I am not too sure what the ammeter will read, it will most likely read zero, but that would not be the stopping potential.

11. Sep 11, 2010

### fluidistic

Ok thanks again.
Now part 2). I know that if I fire photons with $$E=\frac{hc} {\lambda} \approx 4.90819923 \times 10 ^{-19}J$$, the photoelectrons are left with a kinetic energy of $$1.8425 \times 10 ^{-19}J$$. Evidently it seems that 1 photon and not all photons can have the kinetic energy $$1.8425 \times 10 ^{-19}J$$?

That would mean that the work function is simply $$3.06569893 \times 10 ^{-19}J$$. Converting into eV, this gives around $$1.91 eV$$. Unfortunately I see no material with such a property (looking here: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/photoelec.html), which leads me to think that either the exercise didn't want to match the reality or I made some error(s).

What do you say on my answer for 2)?

12. Sep 11, 2010

### rock.freak667

I don't think that you should bother trying to match it to a metal as they did not tell you what metal it was. Your work function should be correct though for whatever 'metal' the question used.

13. Sep 11, 2010

### fluidistic

Ok thank you.
I'll think about how to do part 3). Any problem I have, I post here.

14. Sep 11, 2010

### fluidistic

For part 4), I think it's when the kinetic energy of the photoelectrons is 0. Therefore when $$\frac{hc}{\lambda _{max}}\approx 1.8425 \times 10 ^{-19}J \Rightarrow \lambda _{max} \approx 1078 nm$$ which is in the infrared I believe. Could you confirm my answer?
Now I'll work on 3).

Edit for part 3): Assuming that each photon detach 1 electron, I get that since the energy of 1 photon is about $$4.90 \times 10 ^{-19}J$$ and that the intensity of the beam is $$3 \times 10 ^{-9}\frac{W}{m^2}$$, it means that there're 6112221362 photons hitting a surface of 1 meter squared over a 1 s lapse of time. This is the same number for the photoelectrons emitted.
Am I alright?

Last edited: Sep 11, 2010
15. Sep 11, 2010

### rock.freak667

I think 3 should be what is the minimum wavelength, though your answer is correct. The definition of the work function would lead to the minimum frequency required: Φ=hf0

That looks correct to me.

16. Sep 11, 2010

### fluidistic

Thank you so much for everything, the problem is solved now. However I'd like a quick clarification: when you say "I think 3 should be what is the minimum wavelength,", I don't understand why there would be a minimum wavelength. I understand that a maximum wavelength implies a minimum frequency. To me there's no minimum wavelength, i.e. any photon whose wavelength is smaller than 1078 nm would detach an electron.
Could you clarify what you meant please?

17. Sep 12, 2010

### rl.bhat

1) If you illuminate a photo cathode in a tube without external battery, electrons will be emitted and fill up the space between the two plates. These electrons will have energies ranging from zero to a maximum value. The maximum KE depends on the frequency of the incident photon.The cathode becomes positively charged. The total charge of the system remains constant.
2) If you connect the battery, the electrons will be transferred from anode to cathode to neutralize cathode making anode -ve, which attracts the electrons. Thus establishing the current in the external circuit. As long as the photons are shining on the cathode, the current will flow.
3) If you connect a 10 V battery across the capacitor with reversed polarity, the electric field produced between the plates prevents the electrons coming out of the cathode, in spite of the photons shining on the cathode. The current will flow just like a charging capacitor. And it will stop when pd across the capacitor equals the emf of the battery. In the absence of the photons the capacitor may take longer time to charge fully.
4) K.E. = hc/λ - hc/λmax

Last edited: Sep 12, 2010
18. Sep 12, 2010

### rock.freak667

Sorry, I forgot the inverse relation between 'f' and 'λ'. Minimum 'f' corresponds to λmax.

19. Sep 12, 2010

### fluidistic

Ok thank you very much guys.
2 questions:
1)Will this happen even if I illuminate a photo cathode with photons of the same frequency? Why would the electrons get different KE? Oh maybe because they are at different level in the atoms...
2)I find strange that we didn't use the information that the pico-ammeter couldn't detect any current. Why was it given in the exercise? Just to confuse me or was it important?

20. Sep 12, 2010

### rock.freak667

No that was to tell you that the pd caused the current to reach zero i.e. the pd was in fact the stopping potential required to stop the electrons from being emitted. No emitted electrons = no current on the ammeter.

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