# Homework Help: Question on Photoelectric effect

1. Oct 26, 2004

### stunner5000pt

Light of intensity 2.0 W/ m^2 is incident on an aluminim surface. The wavelength of the light is 160nm. For aluminium the work function is 4.2eV.

Find the kinetic energy of the slowest moving electrons

So then Kmin would be the wavelength at which the cutoff frequency occurs, yes?

e = hf -> e = hc / lambda and E = 4.2eV = hc / lambda then find the cutoff wavelength?

At this point im stumped still... please do help thank you in advance!

Heres another one part b of this question

Fidn the average rate per unit area at which photons strike the aluminium surface

P = E / t = hf / t = hc / lambda t and lambda = 160nm.

I = P / A = hc / lambda t A. I know intensity, h ,c , lambda and all i need is something over t A giving units of per second per sq m

Am i right? input would be greatly appreciated!

2. Oct 26, 2004

### stunner5000pt

I hink i figured out that Kmin would be zero for the slowest electron becuase the sloest electron may not absorb the photon of the highest energy instead it may absorb the photon partially(?)

3. Oct 26, 2004

### ZapperZ

Staff Emeritus
There are no partial absorption. So flush this thought out of your head.

The electrons in the metals are distributed over a range of energy below what is known as the Fermi energy. The most energetic photoelectrons that one obtains came from this Fermi energy level. These are the electrons that had KE of h*nu - phi, the most energetic ones.

However, it doesn't mean that the incoming photons cannot also excite electrons from below the Fermi level. Let's say these electrons are at an energy level epsilon below the Fermi level. Thus, these electrons, when they are emitted, will have an energy h*nu-phi-epsilon, which is now LESS than the most energetic one. The deepest level below the Fermi level that can be emitted will be when h*nu-phi-epsilon = 0.

Zz.

4. Oct 26, 2004

### stunner5000pt

IS epsilon a difference then?
iknow h nu, phi but i dont know epsilon so i can find that out. But how would epsilon lead to kinetic energy??

5. Oct 26, 2004

### ZapperZ

Staff Emeritus
If you don't like "epsilon", call it anything you want. It is just the energy level BELOW the Fermi level, as I've defined earlier.

Zz.

6. Oct 26, 2004

### stunner5000pt

still not getting it

Epsilon (let's leave it like that) is the energy level below the maximum fermi level.

So in order to minimize the Kinetic Energy the value of epsilon must be the largest??

h nu - phi - epsilon = 0 is the least fermi level

for aluminium since phi = 4.2 eV then

h nu - phi - epsilon = (4.14 x 10^-15 )(1.6x 10^-19) c / 160x10^-9 - 4.2 - epsilon = 0

a value of epsilon is found now

so epsion represents the integral difference from the maximum fermi level??

the epsilon i find out is the least kinetic energy that an electron can have asa result , yes?

Last edited: Oct 26, 2004
7. Oct 26, 2004

### stunner5000pt

I hope i actually ahve the correct answwer for the first part.

The second part is the number of photons per second.

8. Oct 27, 2004

### ZapperZ

Staff Emeritus
Here, your calculated "epsilon" represents the DEEPEST level before the top Fermi level that the incoming photon can cause an electron to escape. I answered your question using this because you asked why there is a distribution of energy of the outgoing photoelectrons. I explain that a photon just doesn't excite only the electrons at the topmost Fermi energy state. It can also excite the electrons below that too. Whether these electrons can be excited enough that it will leave the metal depends entirely on the energy of the photon, the work function, and how deep below the Fermi level they are in. I did not realize you had to also calculate this energy level! I simply gave this as a qualitative explanation to make sure you do not hang on to this "partial photon" fallacy.

Epsilon is NOT a measure of the "cut-off" freq. The cut-off frequency is DEFINED as the smallest freq. of the incoming photon that will cause a photoelectron to be emitted. Look at your photoelectric effect equation! If you are given the work function, what is the smallest h*hu to give you the threashold for photoemission? This should be obvious.

Zz.

Last edited: Oct 27, 2004