# Coulomb potential in hydrogen atom nomenclature

1. Jan 15, 2015

### DiracPool

I'm seeing a version of the potential as -Ze^2/4πεr.

My question is what exactly does the Ze^2 refer to? I think the e^2 is supposed to represent the proton and the neutron, and the Z is supposed to represent the number of protons, but I'm not sure how to read it. Does e refer to the charge sign or to the actual value of the charge in coulombs? And Z seems to multiply the whole thing, both the proton and electron together. That doesn't make sense if it's just supposed to represent how many protons there are.

2. Jan 15, 2015

### UncertaintyAjay

Surely you mean proton and electron?
The value of the charge - 1.6*10^-19 C.
Yes it does. Bohr's model, which is what you're referring to applies to single electron species, which includes but isn't restricted to , the hydrogen atom. The force experienced by the single electron will be due to the charge of the nucleus, which is Z times the charge of a proton , which is e( in magnitude). So for hydrogen, plug in Z=1, for He+, plug in Z=2, for Li+2, plug in Z= 3 and so on.

3. Jan 15, 2015

### Staff: Mentor

The Coulomb potential between two charges is proportional to $q_1 q_2$. What are $q_1$ and $q_2$ when you have one nucleus and one electron?

4. Jan 15, 2015

### DiracPool

Yes I did, sorry for the typo.

This may be where I'm having the problem. q is the fundamental charge if I'm not mistaken. So 2 protons would have a Coulomb potential of +2q? Two electrons -2q? One proton and one electron would be zero? I'm not sure. If the value of the fundamental charge, e, is - 1.6*10^-19 C, then wouldn't the value of Ze^2 be that value of e squared multiplied by Z? If Z is simply the numbers of protons, why are we multiplying by that number by e^2, which implies that we are multiplying it by a proton-electron combo? I'm just not getting this..

5. Jan 15, 2015

### Staff: Mentor

Lets settle on a convention: the fundamental charge is $e$, and its value in SI units is 1.6 x 10-19 C.

So, what is the charge $q_1$ of a nucleus with atomic number $Z$ (meaning it has $Z$ protons)? What is the charge $q_2$ of an electron?

6. Jan 15, 2015

### UncertaintyAjay

That's exactly what I'm getting at. The force experienced by an electron in Bohr's model is due to the coulomb force of all the protons in the nucleus. So if you have a system with 1 proton and 1 electron, the coulomb force is e( the charge of the electron) times e( the charge of the proton) times the constant divided by the square if their distance. But suppose you have another atom, say, one with 2 protons and 1 electron( He+), then the force is the the charge of the nucleus( which is Z times e) times the charge of the electron, divided by r^2 blah blah blah. What's important is that your formula refers to atoms that necessarily have one electron but may have more than one proton in the nucleus. In such cases, the charge if the nucleus is given by Z*e.

7. Aug 16, 2015

### Bernhard Kup

May I correct this: the charge of an electron is -e, the charge of a proton ist +e!
Thus the Force between the two is proportional to q1*q2, or (-e)*(+e) = - e2
B.K.

8. Aug 16, 2015

### Bernhard Kup

May I correct this: the charge of an electron is -e, the charge of a proton ist +e!
Thus the Coulomb force between the two e.g. in a H atom is proportional to
q1*q2 = (-e)*(+e) = - e2.
Z then simply is the number of protons in a nucleus of other atoms.
B.K.

9. Aug 16, 2015

### UncertaintyAjay

Yeah. -e. But that doesn't change the point I was trying to make. Which is how you arrive at Ze^2