What is the frequence of radiation that the proton will absorb?

  • Thread starter xinlan
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  • #1
xinlan
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Homework Statement



If protons with magnetic moment mu in the z direction are in a strong magnetic field of magnitude B in the z direction, what is the frequency f of radiation that will be absorbed by the proton as it transitions from parallel to antiparallel states?
Express your answer in terms of mu, B , and Planck's constant h .


Homework Equations





The Attempt at a Solution



E = mu * B
h*f= mu * B
f = (mu * B)h

but I got wrong..
the feedback says that my answer is off by multiplicative factor..

please help me..
thanks..
 

Answers and Replies

  • #2
lzkelley
277
2
you're on the right track.
Its going to take a certain amount of energy to flip the proton, which you can relate to the frequency.
I think, you're going to have to find the Energy (E) required by integrating the work that needs to be done to flip it. The force that the magnetic field exerts on the proton will be B cross mu (i.e. zero force when they're parallel and maximum force when they're perpendicular). express the cross product in terms of some angle between the magnetic moment direction and the B field direction, then integrate over the angle to find the total work.
Work = Force times distance, or more accurately integral of Force dx (dx = differential element of distance, in this case, dtheta).
Does that make sense?
 
  • #3
xinlan
61
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ehm.. why does this has to do with force and work?
the angle between magnetic moment and B field would be zero because both of them are in the z direction.
 
  • #4
lzkelley
277
2
both start in the Z direction, but to get the proton from parallel to antiparallel it has to go from theta = 0 , to theta = pi. the energy required to perform this flip, is the energy difference between the two states.
We know from elementary physics that the change in energy = the work = the integral over force. we know the force, so we can find the change in energy, which leads to the answer.
 
  • #5
xinlan
61
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so f = (mu*B*cos pi)/h ?
 
  • #6
lzkelley
277
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i can't remember if you multiply or divide by h; and also you need to evaluate between 0 and pi, not just at pi. i.e. mu*B (cos 0 - cospi)
 
  • #7
xinlan
61
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I really appreciate that you help me.. but I'm totally confused..
 
  • #8
lzkelley
277
2
so, the force F = mu*B*sin t (where t = the angle theta).
energy E = integral of F dt evaluated from 0 to pi (the initial angle zero, to the final angle pi). Also, like you said, E = hf

E = Int [ mu*B*sin t] dt = mu*b*cos t (but we have to now evaluate t at the 2 end points, 0 and pi - that's how to take an integral).
E = mu*b*[(cos pi) - (cos 0)] so you plug in the end points, and subtract the 1st from the second ---> cos pi - cos 0 = - 2 (and you can ignore the negative sign legitimately, but we shouldn't bother getting into the reasoning)
so E = mu*b*2.
does that make sense?
 

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