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Stern Gerlach Experiment: How wide are the peaks?

  • #1
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Consider the Stern Gerlach experiment, where the oven has a temperature such that the most probable velocity of the silver atoms is 750 m/s. The atoms are collimated by two slits of 0.03 mm width. The magnetic field has a strength of 1 T, a gradient of 5 T/cm, and the length of the magnet is 3.5 cm. The glass plate is located 100 cm behind the magnet.

Q1: How large is the separation of the two beams on the glass plate?
Q2: How wide are the peaks, if the beam is perfectly collimated?

The answer to the first question above is clear. I have no problems with that exercise.

My problem with the second question is that I do not know what is meant by "the peaks". When we look at the picture on the glass plate, we see two curves which meet at two points. It looks similar to the boundary of a human eye. I suppose the wideness of the peaks corresponds to the distance of two points in the picture given at the glass plate. But which one? Can you please help me? I would appreciate your help!

Best wishes
 

Answers and Replies

  • #2
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It would help if you could post an image of the figure (the glass plate). But the "peaks" are the places where the particles of a given angular momentum accumulate.
 
  • #4
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Anyway the peaks are basically the same as you observe in diffraction of light. The width of the peaks corresponds to the width of the slits. Since it says you can assume the beam is perfectly collimated, the variation of the velocity of the atoms perpendicular to the beam doesn't enter into the problem.
 
  • #5
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So the width of the peaks may be due to the width of the slits, but not necessarily "equal" to the width of the slits.
 
  • #6
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Ok thank you for your help. How can I compute the width of the peaks then?
 
  • #7
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Well I am unsure about this. May I ask, what is the level of the course in which this problem is given? Do you think we are expected to take into account the distribution of velocities of the atoms leaving the oven?

One factor that affects the width of the peaks is the variation in the velocity ##v## of the silver atoms. As they pass through the magnetic field, they acquire an impulse ##\int F dt## along the direction of the magnetic field gradient. ##F## is the magnetic force. The impulse would be just ##FT## where ##T## is the time it takes to get through the field. This would vary according to the velocity of the atoms: ##T=L/v## where ##L## is the length of the magnet. The atoms acquire a velocity ##v_p=FT/m=\frac{FL}{mv}## perpendicular to their path, where ##m## is the atomic mass. So the peak width would have a contribution from the variation in ##v##.
 
  • #8
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The problem is given in an undergraduate course (fourth semester).

I am sorry but I think I did not understand it fully. How does the peak explicitly depend on the velocity?
 
  • #9
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Does your course cover the distribution of velocities in a hot gas?

I believe the angle of deflection, in radians, is approximately ##\frac{v_p}{v}=\frac{FL}{mv^2}## so the variation in the angle of deflection (that is, the peak width) would be ##\frac{\Delta(v^2)}{v^2}## where ##\Delta## is the variation in ##v^2##.

I hate to think I am making this solution too complicated. The instructor could be looking for something much simpler than this. That is why I was asking whether the course covers the distribution of velocities.
 
  • #10
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Thank you very much for your help. I think my course does not cover the distribution of velocities in a hot gas. Is there a simpler idea, when we neglect the distribution of velocities?
 
  • #11
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I cannot think of any other factor that would widen the peaks. It says the beam is perfectly collimated, so it seems the peaks should be "perfect."
 

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