Bainbridge Mass Spectrometer - Finding the mass of ions (1 Viewer)

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In a particular spectrometer, doubly ionised C (atomic no. 6 mass no. 12) and singly ionised Li (atomic no. 3 mass no. 6) atoms are
detected. The ratio of the path radii is 1.00252, the Li having the larger value. The
fields are constant.

a)Find the mass of the lithium ion.

Okay so i have already figured out that i have to use the equation:

R=mV/qB (where m = mass, V = velocity, q= charge on electron, R= radius made by the ion and B= magnetic field strength)

I also have the equation:

q(1)R(1)/m(1) = q(2)R(2)/m(2)

(where the numbers in brackets are meant to be subscript)
This equation comes from knowing the value of V/B is constant.

so q(1)= 1.6x10^-19
q(2)= 3.2x10^-19
R(1)/R(2) = 1.00252

(R(1) being the radius of the lithium ion since it is stated to be the largest value)

This takes me to:

m(1)/m(2) = 0.50126

And this is where i get stuck! I cant see a way of doing this question without knowing the mass of carbon. I emailed my lecturer but all he said was 'The mass of carbon is implicit'.
Can anyone help? Thanks! :)
If your professor says that the mass of carbon is implicit, I suppose he means that it is 12.011 amu. This gives you a mass of lithium to be 6.021 amu, which is a little off from the accepted value of 6.941.

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