How Do You Determine the Allowed Radii in a Quantized Angular Momentum Scenario?

[scoldham]

Homework Statement

A particle of charge q and a mass m, moving with a constant speed v, perpendicular to a constant magnetic field B, follows a circular path. If in this case the angular momentum about the center of this circle is quantized so that [tex]mvr_n = 2nh[/tex], determine the allowed radii for the particle in terms of n, h, q, and B for n = 1,2,3,...

Homework Equations

[tex]F = qvBsin\vartheta [/tex]

The Attempt at a Solution

As far as I can tell, this has something to do with relating magnetism to the quantum level. It is easy enough to calculate the radius at a given energy level by solving for [tex]r_n[/tex]. But I do not understand how to relate the charge and the B field to the situation. The best I can come up with is the formula provided... I feel like there is some way it ties into the problem. Help greatly appreciated.

 

[kuruman]

Use the relevant equation you have provided to write Newton's Second Law, F = ma. What is the acceleration for circular motion?

 

[scoldham]

[tex]\frac{mv^2}{r_n} = qVB sin \vartheta [/tex]

[tex]sin \vartheta = 0[/tex] as the angle of the particle with the B field is 90 degrees.

So, simplifying I get,

[tex]r_n = \frac{mv}{qB} [/tex]

How do I tie in this equation with the above?

 

[thebigstar25]

try to use your original equation mvr = 2nh again in the last equation to get red of mv ..

 

[scoldham]

I think I see it now.

[tex]mv = \frac{2nh}{r_n}[/tex]

Subbing mv into equation from above [tex]r_n = \frac{mv}{qB} [/tex]

I get

[tex]r_n = \frac{2nh}{r_nqB}[/tex]

A bit more simplification yields [tex]r_n = \sqrt{\frac{2nh}{qB}}[/tex]

Is that correct?

 

[thebigstar25]

well, it seems correct to me since you achieved what is required in the question which was asking to write r in terms of n, h, q, and B ..