Charges of ions in a mass spectrometer

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SUMMARY

The discussion focuses on calculating the charge of 126C ions in a mass spectrometer using the formula F=qvB=mv²/r. The ions, with a speed of 5.67 x 10^4 m/s and a magnetic field strength of 3.2 x 10^-2 T, yield a charge of 3.2 x 10^-19 C, equivalent to 2e. A point of confusion arises regarding the mass used in the calculation, as the mass of a 126C atom is approximately 1.99264654 x 10^-26 kg, leading to questions about the initial value of 14 atomic units used in the solution.

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  • Knowledge of atomic mass units and conversions
  • Basic proficiency in algebra and physics calculations
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This discussion is beneficial for physics students, particularly those studying mass spectrometry, as well as educators and professionals involved in analytical chemistry and instrumentation.

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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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