Particles in a Magnetic Field Question

[pauladancer]

Homework Statement

A beam of doubly ionized particles (i.e., twice the elementary charge) is accelerated across a potential difference of 2000 V in a mass spectrometer. They are then passed perpendicularly through a magnetic field of 0.085 T resulting in a radius of curvature 12.5 cm. Calculate the mass of the unknown ion.

Homework Equations

Fc= mv²/r
Fm= qvB
ΔV=ΔE/q

The Attempt at a Solution

We are supposed to derive a formula for q/m without velocity. I think the solution probably involves the centripetal force being equal to something but I'm not sure what. Any help would be appreciated, thank you!

 

[gneill]

In your third relevant equation, where does the ΔE end up?

 

[pauladancer]

In your third relevant equation, where does the ΔE end up?

ΔE is the change in potential energy experienced by the particle. I could find that, but I'm not sure what I would do from there.

 

[gneill]

ΔE is the change in potential energy experienced by the particle. I could find that, but I'm not sure what I would do from there.

How is the change in energy expressed by the particle?

 

[pauladancer]

How is the change in energy expressed by the particle?

Sorry, I'm not quite sure what you mean! To be honest I'm not even sure if I need that equation.

 

[gneill]

Sorry, I'm not quite sure what you mean! To be honest I'm not even sure if I need that equation.

Oh, you definitely need it

What happens to a charged particle when it "falls" through a potential difference? What about the particle changes?

 

[pauladancer]

Oh, you definitely need it

What happens to a charged particle when it "falls" through a potential difference? What about the particle changes?

Haha ok good! I guess it gains kinetic energy and loses potential energy, would that ΔE be equal to 1/2mv²?

 

[gneill]

Haha ok good! I guess it gains kinetic energy and loses potential energy, would that ΔE be equal to 1/2mv²?

Why, yes it would!

 

[pauladancer]

Why, yes it would!

Perfect! Ok, so I can say that ΔE= 6.4x10^-16 J which would also equal 1/2mv². After that I'm still not sure what to do!

 

[gneill]

Perfect! Ok, so I can say that ΔE= 6.4x10^-16 J which would also equal 1/2mv². After that I'm still not sure what to do!

You now have an expression involving m and v. What's another relationship that involves m and v once the particle enters the magnetic field?

 

[pauladancer]

Fc= mv²/r ?

 

[gneill]

Fc= mv²/r ?

Try it.

 

[pauladancer]

Try it.

OH! So I can say Vq= 1/2mv² and mv²/r= qvB... So I solved for v on both sides and got v=qBr/m and v=√2Vq/m and then combine those, solve for q/m and so q/m=2V/B²r²! I plugged in the numbers and got the right answer, 9.0x10^-27 kg. Thank you so much for your help, glad I finally got it :)

 

[gneill]

Well done! Glad I could help.