What is the energy difference of a hydrogen atom in a magnetic field?

[knowLittle]

Homework Statement

• Consider a hydrogen atom in its ground level,
placed in a magnetic field of 0.7 T along the z axis.
(a) What is the energy difference between the spin-up
and spin-down states? (b) An experimenter wants to
excite the atom from the lower to the upper state by
sending in photons of the appropriate energy. What
energy is this? What is the wavelength? What kind of
radiation is this? (Visible? UV? etc.)

This is more a theoretical question.
I have received mixed answers from people. Can anyone clarify?

Homework Equations

The Attempt at a Solution

a.) By theory I have read, supposedly there is no spin for s-levels. In other words, In ground state there is no spin-up and down and therefore no splitting of Energies by a magnetic field.
So, There is no energy difference, because there is no spin-up or down on the ground state.

b I guess it would be En=Eo/(n^2)

Help please.

 

[rude man]

The electron can have spins of +/- 1/2.

Use the Sommerfeld expression for the energy level of the atom, with n =1, l = 0, and so j = l +/- 1/2 = +/- 1/2. Then the delta energy between j = +/- 1/2 is your answer.

Photon energy is of course hf.

Hint: use the Rydberg constant to facilitate the math.

 

[knowLittle]

So, in other words. At the ground state, there is still spin of the electrons, but they don't produce a splitting in the energy due to the magnetic field?

Then, a.) would be ΔE= +-1/2 But, +-1/2 what?
(1/ λ)= R( (1/1)- (1/4) ) (Lyman series)
λ=91nm (UV light)

b.) E=h*c/λ

Would this be correct?

 

[janhaa]

In NMR for example, I would say:

\Delta E(proton)=g\mu B_o

g : g-factor
\mu: nuclear magneton
B_o:external magnetic field

 

[knowLittle]

I don't know what NMR or g-factor is.

Can anyone answer my previous post?

 

[rude man]

So, in other words. At the ground state, there is still spin of the electrons, but they don't produce a splitting in the energy due to the magnetic field?

Then, a.) would be ΔE= +-1/2 But, +-1/2 what?
(1/ λ)= R( (1/1)- (1/4) ) (Lyman series)
λ=91nm (UV light)

b.) E=h*c/λ

Would this be correct?

Look at https://en.wikipedia.org/wiki/Hydrogen_atom

for the Sommerfeld energy expression. I think I told you wrong; l = 1 otherwise the expression gives zero as the energy difference between the two electron spin orientations.

So E = R{1 + α²(1/(j + 1/2) - 3/4)}
with R = Rydberg constant ~ 13.6 eV and α the fine structure constant ~ 1/137.


I myself am not sure why (or that) l = 1 in the H ground state. If l = 0 you would be right in that the spin difference gives zero energy difference. Maybe one of our experts can lend a hand.

For b) you have the correct formula.

 

[janhaa]

In NMR for example, I would say:
\Delta E(proton)=g\mu B_og : g-factor\mu: nuclear magnetonB_o:external magnetic field

And in ESR (EPR)

\Delta E(proton)=g\mu_b B_o
g : g-factor
\mu_b: Bohr magneton<br /> B_o:external magnetic field

where \mu_b: is related to the Zeeman effect

 

[knowLittle]

Janhaa, I realize that you are talking about some applied spetroscopy properties, but I should be able to use "basic" concepts of Modern Physics to solve this problem. It's an introductory class to Modern Physics.

 

[janhaa]

\Delta E(electron)=g\mu_b B_o
g : g-factor
\mu_b: Bohr magneton<br /> B_o:external magnetic field

where \mu_b: is related to the Zeeman effect

 

[knowLittle]

Ok, so I have an update on the theory. I misinterpreted it.

Normal Zeeman effect doesn't take into account spin up or down of electrons. So, ground state or s-levels have no energies' splittings.

Anomalous Zeeman effect shows splitting in s-levels:
$\Delta E= \pm \mu _{B} B$
Separation of levels is $= 2 \mu_{B} B$

For part a:
So, about the problem s-levels means that l=0
Energy difference between spin up and down is
$\Delta E= \pm \mu _{B} B = \pm (5.79) 10^{-5} \dfrac {eV}{T} 0.7 T= \pm 4.053 10^{-5} eV$


In part b, I am not sure anymore.
Any help?

 

[knowLittle]

Ok. So, I guess that to excite an atom to an upper level we need
$E_{\gamma}= + 4.053 10^{-5} eV= \dfrac{hc}{\lambda}$
Solving for λ:

$\lambda =\dfrac{hc}{4.053 10^{-5} eV} = \dfrac {1.24 10^{-6} eV m} {4.053 10^{-5} eV}= .0305946213m = 30 594 621.3 nm$

What is wrong with this?

 

[TSny]

For part (a) you are asked to find the energy difference between the spin down and spin up states. So, you need to find the difference between -4.05 x 10^-5 eV and +4.05 x 10^-5 eV.

In part (b) you are asked to find the energy of a photon that will cause the electron to flip from spin down to spin up. Once you have the correct value for this energy, then you can find λ using the method that you have already shown.

 

[knowLittle]

a.) Separation of levels $(4.05- (-4.05))* 10^{-5} eV = 8.10 * 10^{-5} eV$

b.) $\lambda= 15308 * \mu m$

 

[TSny]

I think that's right. Still need to state the type of radiation (i.e., region of the electromagnetic spectrum).

 

[knowLittle]

This is a radio wave. Most likely from a radar.

Note that for part b.), I am using the energy difference. In this way, I am using energy from part a.) to solve the De Broglie relation.

 

[TSny]

OK, radar. Sometimes called "microwaves". See this chart.

[Minor point: The relationship $E = hf = hc/\lambda$ for a photon is due to Einstein rather than de Broglie.]

 

[knowLittle]

I am using this chart:
http://image.tutorvista.com/cms/images/44/spectrum.JPG

Converting my wavelength to cm shows that it's a radio wave.

 

[TSny]

OK. As you can see, different people will divide the spectrum somewhat differently. The chart you are using incorporates "radar" within a broader heading of "radio waves". "Radar" or "microwaves" is more specific.

 

[knowLittle]

I like my chart better.
Thank you so much for your help TSny!