# Modern physics, Bohr model of atom

Gold Member

## Homework Statement

According to Bohr's atomic model,
1)Determine the allowed radius of the orbit of the electrons. Calculate the radius of the first orbit for the hydrogen atom.
2)Show that the energy of the electron is quantized. Calculate the energy corresponding to an electron on the first Bohr's orbit in an atom of H.
3)Justify the use of classical mechanics instead of relativistic mechanics for light atoms.
4)Redo part 1 and 2 assuming that the nucleous' mass isn't infinite.

Somes.

## The Attempt at a Solution

1)I erroneously considered only the H atom. I started by equating coulomb's force to centripetal force. Then writing down the kinetic and potential energy of an electron in the atom.
Then I used the fact that the angular momentum is quantized (I've found this way in a book). Then using some arithmetics tricks, I reached that $$r=\frac{n^2 \hbar ^2}{e^2 k m_e}$$ where k is the constant in Coulomb's law (worth 1 in some unit system, etc.), $$n\in \mathbb{N}- \{ 0 \}$$, $$m_e$$ is the rest mass of the electron and $$e$$ is the electron charge.
So for the hydrogen atom, n=1, right?

If I'm not wrong and Z is the number of protons in the nucleus, I should have reached $$r=\frac{n^2 \hbar ^2}{e^2 k m_e Z}$$ is this right? And in this case, n=2 makes sense while in my first expression it wouldn't, am I right?

2)I reached $$E=\frac{e^4 k m_e}{n^2 \hbar} \left ( \frac{k^2e^2m_e}{2n^2 \hbar ^2} -1 \right)$$, is it right? If I set n=1 I answer the second question of part 2).

3)Should I calculate the velocity of an electron? I could just use the Coulomb's law equated to the centripetal force expression and get v taking a square root. Is this right?
4)Seems really hard.

Related Advanced Physics Homework Help News on Phys.org
So for the hydrogen atom, n=1, right?
No really. The result (r = lengthy stuff) you've got is only for H atom. And in an atom, electron can exist at certain orbits, each of which corresponds to a different value of n. For ground state, n=1. For excited state right above ground state, n=2, and so forth. So any atom, regardless of element, can have any value of n.

To justify your answer, try plugging in the numbers and see if that matches this well-known formula for H atom: $$r = r_0n^2$$ where $$r_0=0.53\times 10^{-10}m$$ (Bohr radius).
(sorry, I don't remember the exact formula; it's too lengthy to me )

If I'm not wrong and Z is the number of protons in the nucleus, I should have reached $$r=\frac{n^2 \hbar ^2}{e^2 k m_e Z}$$ is this right? And in this case, n=2 makes sense while in my first expression it wouldn't, am I right?
Yep, if your answer in (a) is correct, then for an ion which has Z protons & only 1 electron, the formula should be like that (divided by Z). Remember that Bohr model only works approximately well for atoms and ions that have 1 electron, since the model doesn't take into account the interaction between electrons.

Again, n has nothing to do with which element the atom/ion is. The number n shows the quantum behavior of electron. Electron can occupy any value of n depending on the state of the atom/ion (ground / excited state).

2)I reached $$E=\frac{e^4 k m_e}{n^2 \hbar} \left ( \frac{k^2e^2m_e}{2n^2 \hbar ^2} -1 \right)$$, is it right? If I set n=1 I answer the second question of part 2).
I recall that the formula is really lengthy, but not that complicated. For example, for H atom: $$E=-13.6/n^2(eV)$$. Check your calculation 3)Should I calculate the velocity of an electron? I could just use the Coulomb's law equated to the centripetal force expression and get v taking a square root. Is this right?
Yes, in the context of classical view. Bohr model is half quantum (for the quantization of angular momentum part), half classics (for the part pointed out by you ).

4)Seems really hard.
It's the problem of 2 point masses orbiting around their center of mass Gold Member
Thanks for all the valuable info I wasn't aware of, particularly the meaning of n.

I have a question regarding the radius in function of n for heavy atoms. Say you have a ion of gold and that n=3. Does that mean that the extra electron is on the third orbit of the atom?
I know that the first orbit has 2 electrons, the second 8, the third 18, the fourth 32 and so on. So when n increases, does this mean that the extra electron changes of orbit, starting from the first one? Or maybe it could start from any orbit?
And for the H atom, considering n>1 would just mean that the electron is getting farer from the proton.

Yes, if you can remove all the electron but one from Au atom, and given that n=3, that means the electron is on the "third orbit" (strictly speaking, it's on the 3rd energy level).

But that only happens when that electron is the only electron around the nucleus. When you say, n=1 has 2 e- of s orbital, n=2 has 8 e- (2 of s orbital and 6 of p orbital), etc, you are mentioning the quantum result of a NEUTRAL atom, i.e. the atom that is filled with electrons (for Au atom, there are 79 electrons totally). And that's another story, since it's not even Bohr model and not under the constraints of Bohr model.

So for an ion (or H atom exclusively) which has only 1 electron, according to Bohr model, n=1 means the electron is at the n=1 energy level and n=1 orbit, n=2 means the 2nd orbit which is farther than 1st orbit, and so on. Those orbits are NOT what we know as atomic orbital, which is a concept in quantum mechanics, under the constraint of neutral atom. By the way, when we say "atom", we understand that it is neutral atom; otherwise, we should call it "ion".

Gold Member
Ok thank you, I get it.