Download PWE 19-2: Measuring Isotopes with a Mass Spectrometer

Example 19-2 Measuring Isotopes with a Mass Spectrometer
Most oxygen atoms are the isotope 16O (“oxygen-16”), which contains eight electrons, eight protons, and eight neutrons.
The mass of this atom is 16.0 u, where 1 u = 1 atomic mass unit = 1.66 * 10227 kg. The second most common isotope is
18
O (“oxygen-18”), which has two additional neutrons and so is more massive than an atom of 16O: Each atom has a mass
of 18.0 u. To determine the relative abundances of 16O and 18O, you send a beam of singly ionized oxygen atoms (16O+
and 18O+, each with a charge +e = +1.60 * 10219 C) through a mass spectrometer like that shown in Figure 19-8. The
magnetic field strength in both parts of the spectrometer is 0.0800 T, and the electric field strength in the velocity selector
is 4.00 * 103 V>m. (a) What is the speed of the beam that emerges from the velocity selector? (b) How far apart are the
points where the 16O and 18O ions land?
Set Up
We use Equation 19-2 to find the speed v of
ions that pass undeflected through the velocity
selector, and Equation 19-3 to find the radius
of the circular path that each isotope follows.
The distance from where each ion enters the
region of uniform magnetic field to where it
strikes the detector is the diameter (twice the
radius) of the circular path.
Speed of ions emerging from
a velocity selector:
E
(19-2)
v = B
Radius of circular path
followed by charged particles
in a uniform magnetic field:
mv
(19-3)
r =
qB
distance = d18 – d16 = ?
18O
v=?
16O
B
d16
d18
Solve
(a) Calculate the speed of ions that emerge from
the velocity selector. Note that this speed does not
depend on the charge or mass of the ion, so the
16 +
O and 18O+ ions both emerge with this speed.
From Equation 19-2,
v =
4.00 * 103 V>m
E
V
=
= 5.00 * 104 #
B
0.0800 T
T m
From above, 1 T = 1 N # s> 1C # m), so 1 T # m = 1 N # s>C. We also
recall from Chapter 17 that 1 V = 1 J>C = 1 N # m>C. So we can
write the speed as
#
V
C
4 N m
=
5.00
*
10
a
ba # b
T#m
C
N s
= 5.00 * 104 m>s
v = 5.00 * 104
(b) Calculate the radius r and diameter d of the
circular path followed by each isotope.
The masses of the two atoms are
For 16O: m16 = 116.0 u2 11.66 * 10-27 kg>u2 = 2.66 * 10-26 kg
For 18O: m18 = 118.0 u2 11.66 * 10-27 kg>u2 = 2.99 * 10-26 kg
The mass of each positive ion (16O+ and 18O+) is slightly less than the
mass of the neutral atom; the difference is the mass of one electron,
which is 9.11 * 10231 kg = 0.0000911 * 10226 kg. This difference is
so small that we can ignore it.
For 16O,
12.66 * 10-26 kg2 15.00 * 104 m>s2
m16v
=
qB
11.60 * 10-19 C2 10.0800 T2
#
kg m
= 0.104 # #
T C s
r16 =
Since 1 T = 1 N # s> 1C # m2 and 1 N = 1 kg # m>s 2, it follows that
1 T = kg> 1C # s) and 1 T # C # s = 1 kg. So for 16O we have
r16 = 0.104
kg # m
= 0.104 m
kg
d16 = 2r16 = 0.208 m = 20.8 cm
For 18O,
r18 =
12.99 * 10-26 kg2 15.00 * 104 m>s2
m18v
=
qB
11.60 * 10-19 C2 10.0800 T2
= 0.117 m
d18 = 2r18 = 0.234 m = 23.4 cm
The distance between the positions where the
16
O and 18O ions land is the difference between
the two diameters.
Reflect
The distance between where the 16O and 18O ions land is
d18 2 d16 = 23.4 cm 2 20.8 cm = 2.6 cm
This is a substantial distance, so the mass spectrometer does a good
job of separating the two isotopes.
Experiments like these show that, on average, 99.8% of the oxygen atoms are 16O and 0.2% are 18O. (Making up a
small fraction of a percent is a third isotope, 17O.)
Measurements of the ratio of 18O to 16O are important to the science of paleoclimatology, the study of Earth’s
ancient climate. One way to determine the average temperature of the planet in the distant past is to examine ancient
ice deposits in Greenland and Antarctica. These deposits endure for hundreds of thousands of years; deposits near the
surface are more recent, while deeper deposits are older. The ice comes from ocean water that evaporated closer to the
equator and then fell as snow in the far north or far south. Each molecule of water is made up of two hydrogen atoms
and one oxygen atom, which could be 16O or 18O. A water molecule can more easily evaporate if it contains lighter
16
O than if it contains heavier 18O, so the water that evaporated and fell on Greenland and Antarctica as snow contains
an even smaller percentage of 18O than ocean water. This deficiency becomes even more pronounced for colder climates.
It has been shown that a decrease of one part per million of 18O in ice indicates a 1.5°C drop in sea-level air temperature
at the time it originally evaporated from the oceans.
Using mass spectrometers to analyze ancient ice from Greenland and Antarctica, paleoclimatologists have been
able to determine the variation in Earth’s average temperature over the past 160,000 years. They have also analyzed the
amount of atmospheric carbon dioxide (CO2) that was trapped in the ice as it froze. An important result of these studies
is that higher levels of atmospheric CO2 have gone hand-in-hand with elevated temperatures for the last 160,000 years,
which is just what we would expect from our discussion of global warming in Section 14-7. In the same way, the tremendous increase in CO2 levels in the past century due to burning fossil fuels has gone hand-in-hand with recent dramatic
increases in our planet’s average temperature.