質譜儀 - 曲率半徑

The equation describing the radius of the circular path R in a mass spectrometer is:

qvB = mv^2/R
where q is the charge on the ion
m is the mass of the ion
v is its speed when entering into the mass spectrometer
B is the magnetic flux density used
Hence, R = mv/qB ------------------- (1)

Let R1 and R2 be the radius of curvature of CO and N2 respectively
we have, 2(R2) - 2(R1) = 0.65 x 10^-3 m
i.e. R2 - R1 = 3.25x10^-4 m -------------------- (2)

But from (1), R1 = (m1)v/qB where m1 is the mass of CO
(assume v, q and B are the same for the two ions)
R2 = (m2)v/qB where m2 is the mass of N2
thus, R1/R2 = m1/m2 = 28.0106/28.0134 = 0.9999 ---------------- (3)

Solving equations (2) and (3) for R1 and R2 gives R1 = 3.2497 m and R2 = 3.25 m

是否也應該提供質譜儀所有的加速電場及轉彎磁場之強度?