Which formula for the energy of the first orbital of hydrogen is correct?

[physgirl]

Hi,

so I found a difference in what my professor wrote on the board and what I found on wikipedia... my prof said that

E=-Rhc/n^2

but wikipedia says

E=-mq^4/8(h^2)(epsilon)^2(n^2)

So if I write out the expression for R in the professor's equation, it seems that the wikipedia version doesn't include the multiplied term of 1+m_e/m_p in the denominator of the E equation... (m_e is mass of electron, m_p is mass of proton)

Which one is correct?

 

[George Jones]

Which one is correct?

Your prof.

Reduced mass.

Edit: your prof used reduced mass and wikipedia just used the mass of the (edit) electron. Since the mass of the proton is almost 2000 times the mass of the electron, this makes a difference of less than a tenth of a per cent for the energy levels of hydrogen, so some references don't use reduced mass.

 

[physgirl]

What do you mean?

Also, can you explain why you'd factor in the reduced mass?

Lastly, in that case, if you were to calculate the rydberg constant yourself for a hydrogen atom using the wikipedia formula:

http://en.wikipedia.org/wiki/Rydberg_constant

(The first formula they give you)
then you would have to factor in the reduced mass somehow also??

 

[physgirl]

Or wait... I think I might see where the reduced mass is factored in.

So in that case, to find the Rydberg constant myself using wikipedia's equation on that link I just posted, THAT Rydberg equation given on wikipedia already DID take into account the reduced mass, so I'll just plug in numbers for that exact equation to get out R??

But I'm still confused as to when to use reduced mass vs. real mass...

 

[Doc Al]

But I'm still confused as to when to use reduced mass vs. real mass...

Strictly speaking, you should always use reduced mass. But for a hydrogen atom that's a small correction--depending on the specific problem you are solving you can usually ignore it.

 

[George Jones]

What do you mean?

For a single electron atom atom, the reduced mass of the electron is

$$\frac{m}{1+\frac{m}{M}},$$

where $m$ is the mass of the electron and $M$ is the mass of the nucleus (a proton, for hydrogen)

Also, can you explain why you'd factor in the reduced mass?

Because the mass of the nucleus is not infinite. Classically, if the mass of the nucleus is infinite, then the electron orbits the nucleus. If the mass of the nucleus is not infinite, then nucleus bobs around a bit too, and both the electron and the nucleus orbit the centre of mass of the nucleus/electron system. In the centre-of-mass frame, both the nucleus and electron have the same magnitude of momentum $p$, and, for example, the kinetic energy of the system is

$$\frac{p^2}{2M} + \frac{p^2}{2m} = \frac{p^2}{2 \mu},$$

where $\mu$ is the reduced mass. If you have a text for your course, it probably explains things better.

Lastly, in that case, if you were to calculate the rydberg constant yourself for a hydrogen atom using the wikipedia formula:

http://en.wikipedia.org/wiki/Rydberg_constant

(The first formula they give you)
then you would have to factor in the reduced mass somehow also??

Now, I'm a little confused. Isn't this the formula your prof gave?

 

[Shahin]

Or wait... I think I might see where the reduced mass is factored in.

So in that case, to find the Rydberg constant myself using wikipedia's equation on that link I just posted, THAT Rydberg equation given on wikipedia already DID take into account the reduced mass, so I'll just plug in numbers for that exact equation to get out R??

But I'm still confused as to when to use reduced mass vs. real mass...

Strictly speaking you have to use always the reduced mass. When you are talking about the hydrogen atom (in general, when you are talking about the problem of teh two bodies) you can separate the schrodinger´s equation in two terms by defining:

r = r1+r2
MR=m1r1+m2r2

Then you have two independient equations, one for R, and the other one for r. When you solve the equation in r you get de levels for E, so you need to use the reduced mass because it is what you have used in the definiton of your variable´s change. However, in some cases, the diference between using de reduced mass and using de "little mass" is not apreciable and for the sake of simplicity you can use the "little mass".

Sorry for my english.

 

[physgirl]

$$\frac{p^2}{2M} + \frac{p^2}{2m} = \frac{p^2}{2 \mu},$$

where $\mu$ is the reduced mass. If you have a text for your course, it probably explains things better.



Now, I'm a little confused. Isn't this the formula your prof gave?

Wow, first of all, I've never seen reduced mass explained like that, I never really knew where it came from, but that explains a lot

As for the Rydberg constant formula, my professor never gave us a formula for the constant... he just gave us E=-Rhc/n^2... he didn't really define what R was. So what I'm wondering is: is the wikipedia definition of R a proper one that does already take into account the reduced mass? Because I guess for the energy of first orbital of hydrogen, wikipedia didn't take into account the reduced mass...(http://en.wikipedia.org/wiki/Bohr_model)?

http://upload.wikimedia.org/math/5/1/7/517c6ae64fee93b57d8f8550112dfd09.png <--equation wikipedia gives as the equation for energy...