What is the mass of the ion

shimizua

1. The problem statement, all variables and given/known data
A mass spectrometer applies a voltage of 2.00 kilovolts to accelerate a singly charged ion (+e). This ion then enters a region of uniform magnetic field (B= 0.400 T) which then bends the ion into a circular path of radius 0.305 m What is the mass of the ion?


2. Relevant equations
after breaking everything down from F=ma i got to m = (qB^2r^2)/2V


3. The attempt at a solution
well i am just unsure of how to find q. I know that V=2kV, B=.4 T, and r=.305 m. so if you can just help me with finding q. i can do that rest

LowlyPion

Isn't q given as +e?

http://en.wikipedia.org/wiki/Elementary_charge

shimizua

ok well so i tried doing the m= -1.602x10^-19(.4^2)(.305^2)/4 and got -5.96x10^-22. would it be in kg though?

LowlyPion

To put it in perspective for a single proton:

drastichawk

The conundrum you got going here is that the charge is +e, not e. SOO...in taking your big step in finally realizing that q is given as the charge +e, you can now go figure that the charge is POSITIVE (1.6*10^-19 C), not negative.