I Photon Emitted without Changing Energy Levels

[rtareen]

TL;DR Summary

Attached is a problem from my Book (Young & Freedman University Physics 14E), where they are softly claiming that if an electrons orbital quantum number goes down, then it loses energy. This is completely new to me as before we were told that the energy only depends on the principal quantum number.

In Example 41.5, they are implying that, for a hydrogen atom, if the orbital quantum number $l$ goes down the electron will lose energy. However, they said nothing about the principal quantum number $n$ going down, so there should be no loss in energy. As far as this book has presented, the energy only depends on the principal quantum number, but now they are saying it will lose energy if the orbital quantum number goes down. In fact, the energy level equation they gave was

$E_{n} = -\frac{1}{(4\pi \epsilon_0)^2} \frac{m_rZ^2e^4}{2n^2\hbar^2} = -\frac{(13.60~\text{eV})Z^2}{n^2}$

I know its losing energy because that's how the photon is emitted. Can somebody explain this?Also, does Zeeman splitting just mean that for a given $n$, the single energy level becomes split into multiple energy levels given by
$E_n + U(m_l)$?

So if $(n,l) = (2,1)$ then the single energy level is split into three? Particularly

$E_2 + U(-1)$ and $E_2 + U(0)$ and $E_2 + U(1)$

Is that right?

 

[Vanadium 50]

This is completely new to me as before we were told that the energy only depends on the principal quantum number.

Given that the chapter appears to be titled "The Zeeman Effect" it seems very peculiar that they do not mention the changes in energy levels caused by the Zeeman effect. Are you sure this comes out of the blue and is never mentioned at all in the chapter?

 

[PeterDonis]

they said nothing about the principal quantum number $n$ going down

They did not say anything about it not going down either. The problem doesn't even say what kind of atom; it's obviously not hydrogen (can you see why?), but that's all you can deduce about the atom from the problem statement. So you don't know its energy levels.

You can, however, deduce that $n$ does have to change in the transition described. Again, can you see why?

As far as this book has presented, the energy only depends on the principal quantum number

No, that is not correct. At least one statement that contradicts it is on the very page you attached. Read it again, carefully.

does Zeeman splitting just mean that for a given $n$, the single energy level becomes split into multiple energy levels given by
$E_n + U(m_l)$?

So if $(n,l) = (2,1)$ then the single energy level is split into three? Particularly

$E_2 + U(-1)$ and $E_2 + U(0)$ and $E_2 + U(1)$

Is that right?

Yes. Can you see how this answers your other questions?

 

[PeterDonis]

it seems very peculiar that they do not mention the changes in energy levels caused by the Zeeman effect

They do mention it, on the very page the OP attached.

 

[Vanadium 50]

They do mention it, on the very page the OP attached.

You are correct, sir. <sigh>

 

[rtareen]

it's obviously not hydrogen (can you see why?),

I can see now that is not hydrogen. Hydrogen doesn't have any two energy levels whose difference is 2.07 eV. Is that how you saw it too?

You can, however, deduce that n does have to change in the transition described. Again, can you see why?

Unless the answer is the obvious one, I don't see why. I just know that, if the atom is not in a magnetic field, then a change in energy corresponds to a change in n. Is that what you're looking for?

No, that is not correct. At least one statement that contradicts it is on the very page you attached. Read it again, carefully.

If the atom is in a magnetic field, then the energy level depends on both n and $m_l$.

Yes. Can you see how this answers your other questions?

Let me know if my answers are what you're looking for. However, I still believe this is a very bad presentation by the book, because they made it seem that $l$ has something to do with the energy level. This can only be true if $m_l = \pm l$. But they didn't bother to mention that in the problem.

 

[PeterDonis]

Hydrogen doesn't have any two energy levels whose difference is 2.07 eV. Is that how you saw it too?

Yes.

Unless the answer is the obvious one, I don't see why.

You know the difference in energy levels is 2.07 eV. What possible kinds of energy levels could have that much difference in energy? ("Kinds of energy levels" means things like: different $n$, different $l$ values in a magnetic field with the same $n$, etc.)

If the atom is in a magnetic field, then the energy level depends on both n and $m_l$.

Yes.

I still believe this is a very bad presentation by the book, because they made it seem that $l$ has something to do with the energy level.

More precisely, they said $l$ has something to do with the energy level if the atom is in an external magnetic field. At least, that's what I see in what you posted.

This can only be true if $m_l = \pm l$.

Meaning, if $m_l = 0$, then the energy is the same as the $l = 0$ energy level? Yes, that's true for this particular kind of splitting.

But they didn't bother to mention that in the problem.

I would say that's because the textbook authors thought it was simple enough that you should be able to figure it out for yourself. And you did.

Also, saying that $l$ has something to do with the energy level does not imply that every energy level for a given value of $l$ must be different from every energy level for some other value of $l$. It's just saying that you have to know $l$ to know the full set of energy levels. I don't see what the textbook is saying as being misleading in that respect.

 

[rtareen]

You know the difference in energy levels is 2.07 eV. What possible kinds of energy levels could have that much difference in energy? ("Kinds of energy levels" means things like: different n, different l values in a magnetic field with the same n, etc.)

Oh Ok I understand now. Thats too big to just be a change in $m_l$.

 

[PeterDonis]

Thats too big to just be a change in $m_l$.

Yes, you've got it.