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

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    Co2 Curvature
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To separate singly charged CO2 isotopes 12C and 14C, the radius of curvature can be calculated using the formula r = mv/qB. The charge q for the CO2 ion is typically one elementary charge (e) since it is singly charged. The mass of the isotopes will differ, with 12C being lighter than 14C, leading to different radii of curvature in a magnetic field. The discussion clarifies that only the additional charge for ionization is considered, which is one electron charge. Understanding these principles allows for effective separation of the isotopes based on their mass differences.
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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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