In a mass spectrometer a particle with mass m and charge q is first accelerated

Question

In a mass spectrometer a particle with mass m and charge q is first accelerated through a potential difference V. It then enters a region where a uniform magnetic field (B) causes it to move in a circle, with radius r

a) Develop a relation for r in terms of q, m, V, and B.

(b) What is the radius for a proton that is first accelerated through a potential difference of 300 V and then moves in a circle in a magnetic field of 0.05 T?

(c) Use proportional reasoning to find the radius for an alpha particle.

(d) Again using proportional reasoning, by what factor will the radius be different for an electron compared to that for a proton?

Explanation / Answer

Part A)

qV = .5mv^2

v = sqrt(2qV/m)

We also, know that r = mv/qB, thus by substitutuon

r = m/qB (sqrt 2qV/m)

That simplifies to r = 1.41 sqrt(mV/q)/B

Part B)
r = 1.41[sqrt (300)(1.67 X 10^-27)/1.6 X 10^-19)]/.05

r = .050 m (5 cm)

Part C)
r = 1.41[sqrt (300)(6.64 X 10^-27)/(2)(1.6 X 10^-19)]/.05

r = .07 m (7 cm)

Part D)

r = 1.41[sqrt (300)(9.11 X 10^-31)/(1.6 X 10^-19)]/.05

r = .0012 m 1.2 mm

The radius of the proton is 42.9 times larger

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