Radii of Curvature for Separating CO2 with 12C and 14C

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SUMMARY

The discussion focuses on calculating the radii of curvature for singly charged CO2 molecules containing isotopes 12C and 14C. The relevant formula used is r = mv/qB, where m is mass, v is velocity, q is charge, and B is the magnetic field strength. The participant initially miscalculated the charge of CO2, assuming it to be 22e, but clarified that the charge for ionization is typically one electron (1e). This understanding is crucial for accurately determining the radii of curvature in a magnetic field.

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negat1ve
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Homework Statement


Use the same device to separate singly charged CO2 having 12C and 14C. What are the radii of curvature?

This is a follow up question based on a HW problem I answered last week. The original problem was this:

A doubly charged helium atom is accelerated by a voltage 2700V. What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.340-T field?

Homework Equations



r = mv/qB

The Attempt at a Solution



I solved the original problem by saying
q = 2e for the doubly charged He

On accelerating through V volts
1/2mv^2 = qV = (2e)V

Where v = velocity acquired
v = squrt(4eV/m) = 2*squrt(eV/m)

I figured out the mass to be m = 7.26 x 10^-26kg

And then plugging into the formula.

Now this new follow up question is throwing me off a little bit. Based on the question asked and info given how would you separate the CO2? And what would be the charge q of the CO2? I am thinking it would be the 6e from C + 8e(2) of O2 = 22e. Is this correct? Thanks!
 
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No it's only the extra charge you put on to isonise it - generally just one e-
 
mgb_phys said:
No it's only the extra charge you put on to isonise it - generally just one e-

Oh ok I get it now. Thanks!
 

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