# Franck Hertz questions

1. Sep 15, 2004

### JamesJames

I am reading about the Franck Hertz experiment from a text and have some questions about it since I am quite confused...

How do electrons interact with mercury and helium atoms? I hope this is not too stupid a question for you guys...

What has to be the energy of the electron so that the atom can be excited upon colliding with it? (How do you know this?)

How would their result depend on the mean free path of the electron in helium or mercury vapour? The book mentioned a bit about this but I think that I still need some clarification.

What does this mean free path depend on?

I am very curious about these things...hopefully knowing these answers will provide me with a clear idea in my head.

Thanks,
James

Last edited: Sep 15, 2004
2. Sep 16, 2004

### JamesJames

any reply would be great guys.

3. Sep 16, 2004

### ZapperZ

Staff Emeritus
OK, well, since no one is attempting this one, I'll bite....

This is a double-edged sword. If you want the painful details of electron collision with atoms, then I have to refer you to a QED text. However, if you want something superficial, then all you need to do is imagine that an atom is like a billiard ball. An electron carries kinetic energy (KE). Upon collision with an atom, the KE is transfered to the atom. Depending on the KE that is transfered, the atom now is in an excited state.

It depends on the atom. The energy of the electron has to be at least equal to an available transition from one energy state to another. How do I know this? I give the electrons a low enough energy, they will excite nothing.

If the mean free path of the electrons is too short, it does not have sufficient time to again energy from the external field that is accelerating it. Remember that the gas+electrons are in an external potential. The gas is neutral, the electrons are not. After each collision with an atom, an electron loses almost all of its KE. It then needs to be re-accelerated by the external field. If it collides with another atom too soon, it doesn't gain that much of an energy as when compared to the situation where it collides much later in time. The mean free path depends on the temperature of the gas (avg. KE), the density of the gas, and the density of electrons (not a major factor but I thought I include it for completeness sake).

Zz.

4. Sep 16, 2004

### JamesJames

Thanks a lot. I found it to be quite useful. I am still confused about the first question....how do electrons interact with atoms? Could you go into a bit more detail interms of the physics maybe?

Thanks,
James

5. Sep 16, 2004

### ZapperZ

Staff Emeritus
The valence shell of an atom provides a "barrier", or a "shell". An external electron colliding with such a shell transfers energy to such a shell. Depending on how energetic the electron is, it can transfer its energy to a deeper, or core level shells.

Zz.

6. Sep 16, 2004

### JamesJames

Your input has been very clarifying....I do have some more questions if you don' t mind though:

1) Why do you expect current maxima (and/or) minima to be equidistant in the mercury part?

Also I am aware of the fact that with the exception of "going up and down" with the voltage, which is the effect of interaction of electrons with mercury atoms, the current grows with the voltage, due to more electrons being pulled out of the source for higher voltage.

2) HOWEVER, how would the current would behave if there was no mercury vapour in the oven? Why?

7. Sep 16, 2004

### JamesJames

Does the difference between maxima (and/or) minima correspond to the excitation energy? How do you know?

What is the best method to calculate the excitation energy? This is quite difficult for me to understand for some reason.

James

8. Sep 16, 2004

### ZapperZ

Staff Emeritus
Y'know, I must have been brain dead. I should have given you this link early on since this could have answered several of your questions:

http://hyperphysics.phy-astr.gsu.edu/hbase/FrHz.html

As for your question on what would happen if there's no gas in the tube as you increase the potential difference, you'll get Ohm's law relationship!

Zz.

9. Sep 16, 2004

### JamesJames

Thanks, now as for my question
What is the best method to calculate the excitation energy? Here is what I feel:

The values of accelerating voltage where the current drop give the excitation potential. Is this considered to be the energy? Or is there a further relation that allows conversion of Voltage into energy? I would have thought that it is the measured separation of the peaks that corresponds to the excitation energy of the involved mercury transition and not a specific voltage itself...could you clarify this for me please?

I am still unsure as to why it is that one would expect current maxima (and/or) minima to be equidistant for mercury....I don' t know, I just don' t seem to get it.

Finally, how do you know about the Ohm' s Law thing? V = I*R ?

James

Last edited: Sep 16, 2004
10. Sep 16, 2004

### ZapperZ

Staff Emeritus
If an electron is accelerated by a potential difference of 1 V, then it gains an energy of 1 eV. So if there is a minima around let's say 10 V, then you can estimate that the electron that was absorbed had roughly a 10 eV KE, and that this 10 eV was the energy absorbed.

I'm not so sure about this. Maybe at that time, the minima just appear to be equidistance because of the poor resolution of the experiment. There are much more accurate means of determining the mercury absorption lines nowadays.

Becuase I've seen it? :)

In a photoelectric effect experiment, when electrons come out of a cathode, they can be accelerated to the anode. If you are continuously collecting all of the electrons emitted, and you start increasing the potential difference, what you will do in increase the KE of the electrons. You will then see a linear relationship between the applied V, and the photocurrent. This is essentially Ohm's Law.

Zz.

11. Sep 18, 2004

### JamesJames

Thanks for all your help....clears so much up.

One more thing ... I read that a certain minimum energy is required to extract an electron from a metal; this is called the "work function" of the metal. The work function for the anode is higher than that of the cathode, resulting in a so-called "contact-potential difference" between anode and cathode. Its effect is to shift the entire Current vs. Voltage curve.

Why is this the case?

And finally(HOPEFULLY) you said "if there is a minima around let's say 10 V, then you can estimate that the electron that was absorbed had roughly a 10 eV KE, and that this 10 eV was the energy absorbed."

So then the 10 eV would be the excitation energy. I would have thought that the peak would correspond to the EXCITATION energy.

Furthermore, the book says that once the excitation energy has been found, the contact potential difference is given by the difference between this value and the first PEAK.....if peak corresponds to the excitation potential, then this CONTACT potential would yield a positive number. If the dip corresponds to the excitation potential, then the CONTACT POTENTIAL would be negative right?

Last edited: Sep 18, 2004
12. Sep 19, 2004

### JamesJames

Any suggestions ? IT would realy help a lot !

13. Sep 20, 2004

### JamesJames

Anything? I would realy appreciate the help.

14. Sep 20, 2004

### student1938

I' m not very knowledgable on this issue....perhaps someone else can help ...I say excitation energy is the difference between peaks and subtract the potential at the location of the first peak to get the contact potential...WAIT for SOMEONE ELSE to back this up.

student1938

15. Sep 20, 2004

### ZapperZ

Staff Emeritus
I have not commented on your latest set of questions because you are making specific references to a particular text, which you have not cited. I think some of these that you are asking are no longer a "generic" part of the experiment and thus, I have no idea what are the specifics surrounding such a thing. For example, why are the cathode and anode JOINED together to make a contact potential? One typically sees such a thing in a PN junction of doped semiconductors.

Without having a clear picture of what's going on, I cannot, and should not, be making any additional responses, or else I'll simply be blabbering about irrelevant things.

Zz.

16. Sep 20, 2004

### JamesJames

Sorry for the misunderstanding. The book is Experimental Physics by Mellisinos. I guess I have managed to clear up the questions that I asked except for the contact potential one. Why does the contact potential shift the I-V curve?

17. Sep 24, 2004

### dlgoff

Electrons are "slightly" bound in metals and require a little energy to break them free inorder to have a electric current. Remember current is the movement of charge per unit of time (negative charged electronics in this case), so when the atoms begins to absorb the energy from the moving electrons, the electrons slow down which means less electrons moving past a point/second, i.e. less current.

Regards

Don

18. Sep 25, 2004

### JamesJames

Mean Free Path explanation !

I am trying to explain mean free path in the Franck Hertz experiment:

The mean free path of the mercury vapor depends on the temperature of the gas, the pressure of the vapor, the density of the gas and the density of electrons. Mean free path is the distance that electrons travel between collisions with the vapor. The mean free path can be expressed by
lambda= 1/(n*sigma )
where is the cross sectional area of the mercury atom and n is the number density of the electrons. The cross sectional area of the mercury atom is given by
sigma = Pi * R^2
where R is the radius of the mercury atom. The value of n can be determined from constants (that are well known) from the following relationship
P = nkBT
where P is the pressure, kB is Bolztmann’ s constant and T is the temperature.
The mean free path must be small compared to the diameter of the tube to ensure that those electrons which have enough energy to collide inelastically but do not, is minimized. A small mean free path ensures that the electrons experience a large number of collisions inside the tube.

A high concentration of atoms (corresponding to a small mean free path) in the space traversed by the electrons favors the occurrence of inelastic collisions at which the low excited state is produced, for the free path between collisions is shorter and there is a correspondingly smaller chance that the electron can gain sufficient energy between inelastic collisions to a excite higher state. The investigation of such higher levels is more feasible at small gas pressures and with a more refined apparatus. As a result, in the present experiment, we explore only the energy levels 63P since the electrons do not acquire enough energy to excite many of the other levels.

Is this explanation making sense? Also, for low T, we get a high n and so a high mean free path which should be bad right? I thought that high T would be good. Can someone clarify this?

James

19. Sep 26, 2004

### JamesJames

I' m sure someone out there knows if I' m correct. Anything would be really useful to me.