Simple Quantum Mechanics Problem

[student1938]

A ground state Hydrogen atom moves at a speed of 4 km/second through a Stern-Gerlach magnet with a magnetic field gradient of 1000 Gauss/cm. What force does the ground state atom feel? How much does the field gradient accelerate the atom? How much does the field gradient deflect the atom? (The length of the Stern Gerlach magnet is 10 cm - that is the atom experiences the 1000 Gauss/cm field gradient for a distance of 10 cm as it travels through the magnet)

I need some help getting started desparately...any feedback would be greatly appreciated.

 

[student1938]

come on someone must know..I just need something to get started now and I will try any suggestions and come back with more questions if necessary...any input would be great !

 

[spdf13]

Well, here's an idea. Since the atom is in its ground state, you don't have to worry about orbital angular momentum, only its spin. Therefore the force is equal to something like:

F=grad(u.B)

where u is the magnetic dipole which is proportional to S (the spin of the electron) and the "." represents a dot product. So I think this problem is like the classical problem of a magnetic dipole in a magnetic field.

I'm sure the text your using has the equation for the force on an electron in a magnetic field. If your using Griffiths, check out page 163 where they say that F is directly propotional to S.

I hope this might get you started a bit.

 

[student1938]

Yes but how do I use S to get F= ma..it is still a bit confusing. I am not sure where to go next..I figured that out about the ground state but don' t seem to know what wil give me the acceleration or something related to it!

 

[student1938]

The text is by Brehm and Mullin

 

[student1938]

Do I say grad(u.B) = ma ? What is u and what is B?

 

[Doc Al]

The force on the hydrogen atoms is due to the magnetic dipole moment of the electron due to its intrinsic spin. A magnetic dipole in an imhomogeneous magnetic field will feel a force F = μ(∂B/∂z), assuming the field gradient is along the z direction. Since the spin component along any direction of the is quantized (+- 1/2), it turns out that the force on the electron (and thus on the hydrogen atom) equals F = +/- μ_b(∂B/∂z), where μ_b = e(hbar)/(2m).

I'm not familiar with the text you are using, but it should explain this stuff in more detail. If it doesn't, find one that does. You must look this stuff up.

Note: μ stands for dipole moment; B is magnetic field.

 

[student1938]

I thought that u = -grad(u.B) Did you leave out the negative sign on purpose? I know about the Bohr magneton but I mean how do I get u in the the question and how do I manipulate that 1000 Gauss/cm field gradient to get B?

 

[Doc Al]

Originally posted by student1938
I thought that u = -grad(u.B)

No, the force = -grad(u.B).

Did you leave out the negative sign on purpose?

It doesn't matter in the end: you have both components of spin.

I know about the Bohr magneton but I mean how do I get u in the the question and how do I manipulate that 1000 Gauss/cm field gradient to get B?

You don't want B, you want the gradient of B. That's given.

 

[student1938]

IS this the way that it should go then?

F = ¦Ìb(dB/dz) = ma where ¦Ìb = e(hbar)/(2m)

e is a constant(charge of electron)
m is a constant(rest mass of electron)

I would then simply plug into get a right?

And then use one of the basic equations of motion to figure out the displacement...am I correct?

 

[Doc Al]

Originally posted by student1938
IS this the way that it should go then?

...

I would then simply plug into get a right?

And then use one of the basic equations of motion to figure out the displacement...am I correct?

Yes.

 

[student1938]

Ok , now for the equation of motion, I tried d2 = d1 + v1t + (1/2)at^2 but then t is unknown...here' s what I think

In the x direction v is a constant..whatever number it is. I can then say v = d / t where v is given and d is also given...this will be the time required for the particle to travel horizontally while experiencing the deflection. So that allows me to get an expression for t where all the variables comprising t are now known.

Then, in the y direction I would say

y2 = y1 + v1t+(1/2)at^2 where a is calculated using the gradient equation, t is what I figured out from the previous line and v1 is given...does this make sense?

Also one last important question F = -grad(u.B). Can I say that upon opening the bracket that since u is a constant, that F = -u(dB/dz) where d represents the partial derivative? Here u would be a bunch of known constants.

Finally is e = - 1.6 * 10^-19 C ? Then what is m ?

Also why did you only say spin of the electron and not of the proton?

 

[Doc Al]

Originally posted by student1938
y2 = y1 + v1t+(1/2)at^2 where a is calculated using the gradient equation, t is what I figured out from the previous line and v1 is given...does this make sense?

Yes, your method of figuring the dynamics is correct. It's just uniformly accelerated motion, sideways to the original velocity.

Also one last important question F = -grad(u.B). Can I say that upon opening the bracket that since u is a constant, that F = -u(dB/dz) where d represents the partial derivative? Here u would be a bunch of known constants.

Of course you can. Didn't you read my earlier post saying just that?

Originally posted by Doc Al
Since the spin component along any direction of the is quantized (+- 1/2), it turns out that the force on the electron (and thus on the hydrogen atom) equals F = +/- μ_b(∂B/∂z), where μ_b = e(hbar)/(2m).

Originally posted by student1938
Finally is e = - 1.6 * 10^-19 C ? Then what is m ?

Yes, e is the charge on the electron. The "m" in the μ is the mass of the electron. (Look it up.) The "m" in F=ma is the mass of the hydrogen atom.

Also why did you only say spin of the electron and not of the proton?

The magnetic moment is inversly proportional to the mass (see the Bohr magneton), so the proton's magnetic moment can be ignored.

 

[student1938]

Thank you very much..I believe that I have everything that I need for this problem...really appreciate the help even with the questions that I am sure must have been silly...anyways it was very useful help!