Charges of ions in a mass spectrometer

AI Thread Summary
In a mass spectrometer, the charge of 126C ions is calculated using the formula F=qvB=mv²/r, resulting in a charge of approximately 3.2*10^-19 C, equivalent to 2e. There is confusion regarding the mass used in the calculation, as the mass of a 126C atom is approximately 1.99264654*10^-26 kg, or about 12 atomic units, rather than the 14 atomic units referenced in the solution. This discrepancy suggests a possible error in the problem statement or a mix-up with nitrogen atoms. Participants in the discussion are trying to clarify the origin of the mass value used in the solution. Accurate understanding of ion mass and charge is crucial for proper mass spectrometry analysis.
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Homework Statement


In a mass spectrometer, a few 126C ions are deflected with a radius of 12.9 cm. The ions have a speed of 5.67*10^4 m/s when they enter the field. The magnetic field has a strength of 3.2*10^-2 T. The charge of these ions must be ...

SOLUTION:

We use the formula: i F=qvB = mv2/r and find q.

q= mv / rB = (14*1,66·10^-27 kg * 5,67*10^4 m/s) / (0,129m * 3,2*10^-2 T) = 3,2*10^-19 C = 2e

Homework Equations



One atomic unit is 1,66*10^-27 kg

But isn't the mass of 126C atom is ca 1.99264654*10^-26 kg, which is ca 12 atomic units? ( https://www.britannica.com/science/atomic-mass )

Where does the 14 come from in the solution?

The Attempt at a Solution

 
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Perhaps the question was modified from Nitrogen atoms to Carbon atoms at some point but the solution was not updated to reflect this?
 
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