# photoelectric effect

A 1.0 mW laser ($\lambda$ = 590 nm) shines on a cesium photocathode (ϕ = 1.95 eV). Assume an efficiency of $10^{-5}$ for producing photoelectrons (that is, 1 photoelectron is produced for every 10^5 incident photons) and determine the photoelectric current.

So the way I tackled this problem was the following:

we know $$KE_{max} = hf - \phi$$ $$KE_{max} = 0.1180 eV$$

now I dont know how to finish? I know the current is a rate of time but there is no time given here? Also what is the stopping potential? I did a little bit of research online and found this:

$$V_{0} = \frac{W}{q}$$ and

$$KE_{max} = eV_{0}$$

I tried equation them but I dont know what to solve for (i solved for e which was the wrong answer).

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Is this a homework question, or something for a class assignment? If so, please add the homework tag (and if not, you should reword it to something more general). – David Z Nov 19 '10 at 5:29
its a practice question from an assignment. Should I mark it as homework? – masfenix Nov 19 '10 at 5:32
HINT: From power you can get number of electrons emitted per second. $P=nE/t$ and $E = hc/\lambda$. – Pratik Deoghare Nov 19 '10 at 6:48

Describe the process at a high level:

• The laser produces a certain number of photons per second $N_{\gamma}$
• The photons free a certain number of electrons per second $N_e$
• The free electrons per second give you the current $I$

Now

$N_{\gamma}= \frac{P_{laser}}{E_{\gamma}}$, where $E_{\gamma} = h\nu = \frac{hc}{\lambda}$

Therefore $N_{\gamma}= \frac{\lambda}{hc}P_{laser}$

$N_e = \eta N_{\gamma}$, where $\eta$ is the efficiency

Therefore $N_e = \frac{\eta \lambda}{hc}P_{laser}$

$I = N_e q = \frac{q \eta \lambda}{hc}P_{laser}$

In other words, the current is proportional to the intensity (power) of the laser as expected:

$I = kP_{laser}$, where $k=\frac{q\eta\lambda}{hc}$

The last thing to check is the cutoff point, in other words if the energy of a photon is enough to overcome the binding energy of the electrons. This you already did.

See here http://physics.info/photoelectric/, in particular where it says

The rate (n/t) at which photoelectrons (with charge e) are emitted from a photoemissive surface can be determined from the photoelectric current (I)...

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Your quantities feel a little weird, there's a typo in your $E_{phi}$ and it's a bit awkward that you redesignate $\Phi$ from the question, but you are correct, so I'll +1 this. – Thomas Themel Nov 19 '10 at 12:49
I'll change the letters, I did this before my morning coffee... ;-) I fixed the typo as well – Sklivvz Nov 19 '10 at 13:31
sweet thankyou. this explained a lot more then the question. I understood the entire chapter because of this lol. – masfenix Nov 19 '10 at 21:47

Note the power units given for the laser intensity. Power is energy transferred or transformed per unit time.

So, how many photons per second, and from that how many photo-electrons?

And an interesting question to ask yourself then is if the metal is electrically isolated, can this go one forever, and if not why and how does it stop? What would you need to know to give a quantitative answer?

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The first thing you have to calculate is the number of photon incident on the photocathode per unit time!

That you can calculate by: $P_{laser}=E_{inc}/dt=[J/sec]$ since every photon carries an energy equal to $h\nu=h\cdot c/\lambda=3.37\cdot 10^{-19} J$ (greater than the work power of course, if not what the heck are we doing here??) you can obtain: $$\frac{P_{laser}}{E_{photon}}=\frac{n}{dt}$$ being $n/dt$ the photon/second flux.

You do the math and you find: $n/dt\approx3\cdot 10^{11}$ photon per second.. (check that!)

The efficiency is $10^{-5}$ so that means that you produce $\approx 3\cdot 10^6$ electon per second: and that is you photoelectric current that you can convert in your favorite unit.

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