Solving for Radius in Magnetic Field: q of Sr+?

AI Thread Summary
The discussion centers on calculating the radius of a magnetic field for Strontium ions (Sr+) using the Lorentz Force equation. The user derived the radius formula as r = (2mV/qB)^(1/2) but is uncertain about how to determine the charge (q) of Sr+. It is clarified that for Sr+, which has one less electron, the charge q is equal to the elementary charge (e). The user acknowledges the guidance and recognizes the need to recheck their derivation for accuracy. Understanding the charge of ions is crucial for solving the problem correctly.
wjdgone

Homework Statement


The Lorentz Force can be used to sort ions (atomic or molecular) based on their charge to mass ratio. This configuration has been used to separate isotopes and as a mass spectrometer. A beam of Strontium ions Sr+ is accelerated through a potential of 500 V and injected into a region of uniform 1 T magnetic field pointed out of the page. Strontium has an atomic mass of approximately 87 AMU (AMU = 1.66 10−27 kg ). Find R in cm.

Homework Equations


r = mv/qB (vector arrows where necessary)
E = qV = (1/2)mv^2

The Attempt at a Solution


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I solved for v with the second equation and plugged that value in the first equation and got r = (2mV/qB)^1/2, which I'm fairly certain that I'm supposed to get. What I don't get is how to solve for q (nor could I find a value online). Everything else is given; is there an equation I'm missing or something? Is q=e because Sr+ has one less electron? It's been a long time since I've done chemistry haha..
 
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wjdgone said:
got r = (2mV/qB)^1/2
This isn't quite right. Check your derivation.

Is q=e because Sr+ has one less electron?
Yes.
 
TSny said:
This isn't quite right. Check your derivation.

Ah, I see. Thanks for your help.
 
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