Particles in a Magnetic Field In a mass spectrograph, ions with a mass m and cha
ID: 2178321 • Letter: P
Question
Particles in a Magnetic Field In a mass spectrograph, ions with a mass m and charge q are accelerated through a potential difference V. They then enter a uniform magnetic field B that is perpendicular to their velocity and are deflected into a semicircular orbit of radius R. A detector measures where the particles exit the semicircle, thereby providing a measurement of R. Derive an equation for calculating the mass m of the ion from measurements of B, V, R, and q. What potential V would be required so that singly ionized 12C ions (i.e. with charge q = e) will execute orbits of radius R = 50.0 cm in a magnetic field of B = 0.150 T? (Take the mass of a 12C ion to be m12 = 1.993(10)-26 kg.) Suppose that the ion beam consists of a mixture of 12C and 14C ions. If V and B have the same values as in (b), calculate the spatial separation of these two isotopes at the detector. (Take the mass of a 14C ion to be m14 = 2.325(10)-26 kg.)Explanation / Answer
accelerate the ions... qV = (1/2) mv^2 so v = sqrt(2qV/m)
magnetic force on ions F = qvB
circular motion F = mv^2 /R
Put these together to get the answer to part a:
(a) F = 2 qV / R and F = q * sqrt(2qV/m) * B
set expressions equal
2qV / R = q * sqrt(2qV/m) * B
simplify
2V/BR = sqrt(2qV/m)
4 V^2 / R^2 B^2 = 2 qV / m
solve for m
m = q R^2 B^2 / 2 V (note: denominator has R squared and B squared)
(b) Solve this expression for V
V = q R^2 B^2 / 2m = 1.60x10^-19 * 0.50^2 * 0.150^2 / 2*1.993x10^-26 =
= 22580 Volts
(c) Note that in the expression, mass is proportional to R^2.
We know that R for the C12 is 50.0 cm. So we can write a proportion:
mass of C14 / mass of C12 = (R for C14 / R for C12)^2
2.325 / 1.993 = (R for C14 / 50.0)^2
R for C14 = 54.004 cm
Since the particles execute a semicircle, their separation will be the difference in their diameters, or twice the difference of their radii. So
separation = 2 * (54.00 - 50.00 ) = 8.00 cm