PHYS 203 Homework #12
Due June 2, 2010
Preliminary comments.
I. Your handout is from a textbook by Cutnell and Johnson. They give masses in atomic
mass units (u), but often do not give enough significant digits. In calculating masses, it
is often better to use the mass excess or mass defect, which is defined as follows. We are
actually using mass energy here; we are expressing mc2 instead of m. One amu has a mass
energy of 931.494 MeV. The mass (actually mass energy) of a nuclide is
((number of nucleons)(931.494 MeV) + (mass excess)
For example, if you look up the mass excess of tritium (31 H) at a website such as
http://www.nndc.bnl.gov/chart/, you find it to be 14.9498 MeV. That means that the
mass energy is 3(931.494 MeV) + 14.9498 MeV = 2809.4318 MeV. (This particular web site
denotes the mass excess as ∆(MeV), which is accurate but they don’t put an explanation
with the label.)
The textbook does an example of the binding energy of the helium nucleus. They do it
using atomic mass units. Let’s do it instead with the mass excess numbers.
• The mass excess of helium-4 is 2.4249 MeV.
• The mass excess of hydrogen is 7.2890 MeV.
• The mass excess of a neutron is 8.0713 MeV.
Therefore the separate masses, before fusion, have a mass excess of
2(7.2890 MeV) + 2(8.0713 MeV) = 30.7206 MeV
The binding energy is therefore 30.7206 MeV - 2.4249 MeV, or 28.2957 MeV., good to
this many digits. Cutnell rounds this off to 3 significant figures, which is fine for many
applications. However, let us calculate the mass defect, which the textbook gives as 0.0304
u. We obtain
28.2957 MeV
= 0.030377 u
931.494 MeV/u
This may seem a trivial difference, but I have found that when you are using atomic mass
units to calculate mass defects, 3 significant digits is not enough.
A comment on significant digits. In the number “4.0330 u” there are only
three significant digits. The leading digit is not significant; it is the exact number
of nucleons. The zero after the decimal point is not significant, as you know. That
leaves 3 digits.
As another example, let us calculate the mass in amu of tritium, whose mass
energy we found earlier. We may write
3(931.494 MeV) + 14.9498 MeV
931.494 MeV/u
14.9498 MeV
3(931.494 MeV)
+
=
931.494 MeV/u
931.494 MeV/u
= 3 (exactly) + 0.016049 u
= 3.016049 u
m =
It should be clear from this example why the “3” is not significant. In the final
number we have only kept 5 significant digits. We could have kept 6, since we
are dividing two numbers which each contain 6 significant digits.
II.In the textbook’s treatment of beta-decay (Chapter 31), you are not quite told the whole
story. For example, an isolated neutron is not stable, and will decay into a proton and
a β − -particle (an electron) with a half-life of about 12 minutes. Beta-decay is, of course,
observed in many other particles. Historically, analysis of beta-decaying particles always
showed that there was some “missing” momentum. Physicists were at one time tempted to
give up the law of momentum conservation! This was finally resolved by Fermi’s postulate of
the emission of a third particle, the neutrino. These are very non-interactive and of zero (or
near-zero) mass. Neutrinos were finally detected in 1956. The complete beta-decay reaction
for an isolated neutron is then
1
0n
−→
1
1p
+ β− + ν
where ν is an antineutrino. (Technically, it’s an electron antineutrino, to differentiate it from
muon and tau neutrinos.) Why is it an antineutrino and not a neutrino? Well, a conservation
law has been observed, that whenever an ordinary particle is produced, an antiparticle must
also be produced, and vice versa. (One might argue that this conservation “law” is invented.
If we express our observations in terms of this conservation law, it gives a pleasing simplicity
and symmetry to our observations, so we eventually think of this invented formalism as a
fundamental law.)
1. (8 points) Carbon-14 has a half-life of 5715 y. It decays via β decay, emitting an
electron.
(A) Determine the rate constant λ of
14
6 C,
in units of y−1 .
(B) Carbon-14 is produced in the upper atmosphere. Cosmic rays generate neutrons,
which are then absorbed by nitrogen nuclei, in the following reaction:
1
14
1
n +14
7 N −→6 C + H
The 14 C then mixes with the atmospheric carbon, so we assume there is an equilibrium established so that the ratio of 14 C to 12
6 C is constant. A “standard” has been
established: it is assumed that at this equilibrium, one gram of carbon will produce
14 disintegrations (β-decay) per minute. Find the number of 14
6 C atoms required to
produce 14 counts per minute. Then determine the total number of carbon-12 atoms
12
in one gram of carbon. Finally find the ratio of 14
6 C to 6 C for these conditions.
(C) We assume that living organisms maintain an equilibrium with the carbon in the
12
atmosphere, so that when, say, a plant dies, it has the ratio of 14
6 C to 6 C that you
found above. Over a long time, the amount of 14
6 C decreases. Suppose we measure the
beta emissions from a piece of charcoal from an archeological dig. We find that for
each gram of carbon, we obtain 5.5 counts per minute. (This is more difficult than just
pointing a Geiger counter at the material. We are trying to catch every single beta
particle emitted in all directions.) Determine the ago of the sample of charcoal.
12
We should note that the assumption of a constant ratio of 14
6 C to 6 C in the
atmosphere is not always true. Nuclear bomb tests have changed this ratio,
at times almost doubling the amount of 14
6 C. One can determine someone’s
14
age by measuring the 6 C in tooth enamel and comparing with the known
amount of 14
6 C in the atmosphere as a function of time!
2. (5 points) Your instructor used to use 32
15 P in enzyme reaction rate measurements.
Typically, the supplier would give you the activity (in Curies) of a sample at some
date and hour. From there, you have to find the activity of the sample you have in
hand. Suppose you receive a shipment of 32
15 P-labeled ATP which was stated to have an
activity of 7.85 µCi per ml, 3.5 days before you receive it. 32
15 P has a half-life of 14.28
days. What is the activity (in µCi/ml of your material? (You don’t have to convert
Curies to any other unit; just use the exponential decay law.)
3. (5 points) Radium-223 has a mass of 223.019733u. It decays by α-decay to Radon-219,
which has a mass of 219.009478u. An alpha particle has a mass of 4.001503 amu. Find
the kinetic energy of the alpha particle, assuming that all the energy due to change in
mass is reflected in the alpha particle’s kinetic energy.
EXTRA CREDIT (5 points). This is not quite true, since the Radon atom
will recoil. Use conservation of momentum to find the kinetic energy of each
particle after the decay.
4. (6 points) Complete the following nuclear reaction equations.
• Decay of 83 Li by β − decay.
• Decay of
• Decay of
• Decay of
• Decay of
• Decay of
15
8 O by positron emission
32
−
15 P by β decay.
97
43 Tc by electron capture.
212
86 Rn by alpha emission.
230
90 Th by alpha emission.
(β + decay).
5. (6 points) (A) Efforts are underway to produce energy commercially by fusion. It is
too difficult to produce helium by fusion of protons, as is done in the sun, so we have
to start with heavier nuclei such as deuterons. One possible reaction is
2 21 H −→31 H +11 H
That is, two deuterium atoms produce one tritium and one ordinary hydrogen. (This
is not a single-step reaction, but that is the result.)
(A) Calculate the energy, in MeV, released in this reaction. (4.033 MeV)
(B) The abundance of deuterium relative to ordinary hydrogen is 0.015%. This means
that in one kg of water, there is 0.0167 mol (or 0.0333 gram) of deuterium. Calculate
the energy available from using all of the deuterium in one metric ton (1000 kg, often
called a tonne) of water for a reaction as the one above. (More energy than this would
be available, since the tritium produced is unstable and will eventually decay to He-3,
with release of energy). Keep two significant digits. (3.2×1012 J)
(C) Bonneville Dam is capable of generating slightly over 500 MW of power. For
what period of time could a fusion-based power plant generate this much power using
deuterium from 1 tonne of water? Assume that 65% of the heat generated by the
nuclear reaction is turned into electricity. (This is about the best we can do with
commercial gas-fired plants.)