Mass spectrometer A mass spectrometer is a tool used to determine accurately the
ID: 1490251 • Letter: M
Question
Mass spectrometer
A mass spectrometer is a tool used to determine accurately the mass of individual ionized atoms or molecules, or to separate atoms or molecules that have similar but slightly different masses. For example, you can deduce the age of a small sample of cloth from an ancient tomb, by using a mass spectrometer to determine the relative abundances of carbon-14 (whose nucleus contains 6 protons and 8 neutrons) and carbon-12 (the most common isotope, whose nucleus contains 6 protons and 6 neutrons). In organic material the ratio of 14C to 12C depends on how old the material is, which is the basis for "carbon-14 dating." 14C is continually produced in the upper atmosphere by nuclear reactions caused by "cosmic rays" (high-energy charged particles from outer space, mainly protons), and 14C is radioactive with a half-life of 5700 years. When a cotton plant is growing, some of the CO2 it extracts from the air to build tissue contains 14C which has diffused down from the upper atmosphere. But after the cotton has been harvested there is no further intake of 14C from the air, and the cosmic rays that create 14C in the upper atmosphere can't penetrate the atmosphere and reach the cloth. So the amount of 14C in cotton cloth continually decreases with time, while the amount of non-radioactive 12C remains constant.
Here is a particular kind of mass spectrometer (see the figure). Carbon from the sample is ionized in the ion source at the left. The resulting singly ionized 12C+ and 14C+ ions have negligibly small initial velocities (and can be considered to be at rest). They are accelerated through the potential difference V1. They then enter a region where the magnetic field has a fixed magnitude B = 0.14 T. The ions pass through electric deflection plates that are 1 cm apart and have a potential difference V2 that is adjusted so that the electric deflection and the magnetic deflection cancel each other for a particular isotope: one isotope goes straight through, and the other isotope is deflected and misses the entrance to the next section of the spectrometer. The distance from the entrance to the fixed ion detector is a distance of w = 28 cm. There are controls that let you vary the accelerating potential V1 and the deflection potential V2 in order that only 12C+ or 14C+ ions go all the way through the system and reach the detector. You count each kind of ion for fixed times and thus determine the relative abundances. The various deflections insure that you count only the desired type of ion for a particular setting of the two voltages.
(b) Determine the appropriate numerical values of V1 and V2 for 12C. Carry out your intermediate calculations algebraically, so that you can use the algebraic results in the next part.
V1 = _____ V
V2 = _______ V
(c) Determine the appropriate numerical values of V1 and V2 for 14C.
V1 = _______ V
V2 = _______ V
Explanation / Answer
a) Accelerating plates: Left is + and right is – so that the charges are accelerated across the gap. Deflection plates: Upper is + and lower is – so that the electric field (and the electric force on a positively charged particle) points vertically down and points opposite to the magnetic force. The positively charged particle bends upwards in the magnetic field, by the RHR the magnetic field has to point into the page.
b) To determine V1, work is done accelerating charge q across the accelerating region.
q* V1 = 0.5mv^2 ====> V1 = mv^2/2q
where the velocity is determined from the bending of the charged particle in the far right external magnetic field. The velocity is thus
FB = qvB = mv^2/R =====> v = qRB/m
Thus we get, V1 = mv^2/2q = (qR^2*B^2)/2m
To determine V2 we equate the electric and magnetic forces in the deflection region. We have
FB – FE = 0 ===> FB = FE =======> qvB = qE = q (V2)/L ===> V2 = BLv = (qRLB^2)/m
Now For 12C
V1 = (qR^2*B^2)/2m = (1.6*10^-19*0.14^2*0.14^2)/(2*12*1.67*10^-27) = 1533.6 V = 1.53 kV
V2 = (qRLB^2)/m = (1.6*10^-19*0.14*0.01*0.14^2)/(12*1.67*10^-27)= 219.1 V = 0.22 kV
Now For 14C
V1 = (qR^2*B^2)/2m = (1.6*10^-19*0.14^2*0.14^2)/(2*14*1.67*10^-27) = 1314.5 V = 1.31 kV
V2 = (qRLB^2)/m = (1.6*10^-19*0.14*0.01*0.14^2)/(14*1.67*10^-27)= 187.8 V = 0.19 kV