Physics 121, Fall 2016
Donald A. Smith
Solutions Set #8: Radioactive Decay, Diffusion,
and Blackbody Radiation
1. An old-fashioned silver quarter was irradiated with slow neutrons. These neutrons are captured by the silver nuclei in the quarter. The quarter was then placed close to a Geiger counter
and the following count rates (in counts/sec) were recorded as a function of time.
a) Using the table of nuclides that has a link on the class web page, determine the reactions
that could happen if the silver nuclei in the quarter absorbed a neutron. Are the products of the
neutron capture stable?
According to the web page, silver exists in nature in roughly equal abundances of both Ag107
and Ag109 . So we can hypothesize that a random sample of silver will contain roughly equal
amounts of each isotope. We will be able to test this hypothesis. If these nuclei were to absorb
a neutron, they would become Ag108 and Ag110 . With a half-life of 2.37 min, 97% of the Ag108
will undergo beta decay to Cd108 , but more than 99% of the Ag110 beta-decays to Cd110 with a
half-life of 24.6 s! That’s not stable.
b) Determine the half-life of the decay of the silver nuclides produced when the silver in the
quarter absorbed a neutron. Use the data to determine your answer. After you have your
answers, look up the half-lives on the table of nuclides on the web and compare.
I entered the table into Graphical Analysis and plotted the natural log of the counts vs. time.
The resulting graph is shown in Figure 1. I estimated the uncertainties by taking the square root
of the number of counts, which I deduced by assuming all activity measurements were averaged
over 20 s intervals, and then dividing by 20 s to get back into activity units.
There are clearly two distinct decays going on. One decay has a much shorter half-life than
the other, so by the time 200 s have gone by, whatever was undergoing the first decay is pretty
much gone, but the longer-lived decay is still happening. So far, this is consistent with the
hypothesis that we are seeing the decay of a mix of Ag-108 and Ag-110, but we need to dig
deeper.
I therefore fit a straight line to the latter part of the curve to get the long-lived decay by
itself. This yields a best-fit line of
y = (−0.0049 ± 0.0001 s−1 )t − 0.74 ± 0.05.
(1)
I can convert Equation 1 back into activity by putting both sides in the exponent of e. Then
I subtracted this exponential curve from all the measured activity (which presumably contains
counts from both isotopes) to get the short-lived decay by itself. I see no evidence for the curve
flattening out at the end, so I assume that any background from the environment is so small
that I can safely ignore it. Then I take the log of that difference to get the straight line. This
result is shown in Figure 2. I have propagated the uncertainties through the algebra using the
standard formula.
When I fit the first nine points in Figure 2 (the points before 200 s, when the error bars are
still reasonably small) to a straight line, I got
y = (−0.031 ± 0.002 s−1 )t − 0.1 ± 0.2.
1
(2)
Physics 121, Fall 2016
Donald A. Smith
Figure 1: Natural log of Activity (normalized) vs. time (in seconds) from an irradiated silver
quarter. Note that there are two straight line segments, with a crossover time of about 200 s.
To get the half-life of each decay, I divide the natural log of 2 by the best-fit slope, and get
140 ± 3 s and 23 ± 1 s. These values are not significantly different from the half-lives reported in
the table (2.37 min is 142 s, which is less than one sigma from my result, and 24.6 s is less than
two sigma from my result), so I conclude that my hypothesis is not wrong: I have observed the
decay of Ag108 and Ag110 .
An alternate way to check this (if you don’t want to have to calculate uncertainties for your
measured half-lives) is to convert the half-lives given in the web chart into predictions for what
the slope of the semilog graph would be, if the hypothesis is not wrong. If you do that, you get
0.00487445 s−1 and 0.02818 s−1 . As before, the slower decay is not different to less than one
sigma, and the faster decay is not different to less than two sigma. Either way, the hypothesis is
not wrong.
2. In the spirit of the season... let’s model the behavior of trick-or-treating kids as a random
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Physics 121, Fall 2016
Donald A. Smith
Figure 2: Natural log of Activity (normalized) vs. time (in seconds) from an irradiated silver quarter. The best fit function for the long-lived activity we attribute to Ag 108 has been
subtracted from the total. Note that after about 200 s, there are no significant (different from
zero) detections. The first 200 s of data therefore represent the activity we can attribute to the
fast-decaying isotope (Ag 110), if the hypothesis is not wrong.
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Physics 121, Fall 2016
Donald A. Smith
walk in two dimensions and perform Fermi problem estimation as to how long it would take them
to cover a whole neighborhood.
a) If we consider a “mean free path” of a child on halloween to be the typical distance between
houses, estimate the diffusion constant for kids on Halloween.
We know that the diffusion constant is
1 λ2
1
,
D = λv̄ =
2
2 ∆t
(3)
where λ is the mean free path and ∆t is the time between interactions. So, I will estimate 20 m
as the distance between houses, and I will estimate that it takes a horde of excited children about
40 s to get from one house to the next (including the time spent at each house).
That yields D ≈ 5 m2 /s.
b) If all the kids are released from a single bus at the center of a neighborhood that extends
roughly 1000 m in any direction, how long would you reasonably expect to have to wait for
the kids to cover the whole neighborhood? Assume that the kids take off from each house in a
completely random direction, not just on a perpendicular grid.
We use < r2 >= 2Dt (I don’t need to include the number of dimensions because I am not
using a perpendicular grid) and get that
t=
10002
s = 105 s.
2×5
(4)
That’s 28 hours – more than a day. i think real kids will be much more efficient than that!
3. The density of thick fog is about 0.05 g/m3 . On a really foggy day, you can see maybe about
20 m. Given these numbers, what would you estimate the opacity of thick fog to be?
Typically, you can see out to where the optical depth is about one. Which means essentially
one mean free path is about 20 m. The opacity would therefore be
τ = κρr → κ =
1
1
=
m2 /g,
ρr
0.05 × 20
(5)
which works out to κ = 1 m2 /g.
4. When we look out into space at sources of light in our galaxy, the light we see has to pass
through the interstellar medium to get to us. If you look at an image of the Milky Way in
optical light, there are large swaths that are dark. On average, there is about one particle per
cubic meter in the interstellar medium. If the scattering in the interstellar medium were due
to Thompson Scattering, what would the mean free path of light in the Galaxy be? Is that
reasonable?
The mean free path can be estimated as ` = (σn)−1 . If we assume Thompson Scattering, then
σ = 6.65 × 10−29 m2 , and we are given that the number density is n = 1 m−3 . Put this together
to get ` ≈ 1028 m. This is about a trillion light years, much larger than the size of the visible
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Physics 121, Fall 2016
Donald A. Smith
universe! How is this possible? Well, the obvious conclusion is that the dominant interaction of
light and matter in the interstellar space in the Galaxy is NOT Thompson scattering. Remember
that Thompson scattering is for photons hitting free electrons. The sun is a hot, dense plasma,
so there are lots of free electrons around. In deep space, it’s way too cold to get many free
electrons. In fact, the dark zones in the Milky Way are caused by visible light hitting grains of
dust, which then move to a slightly higher temperature and glow in the infra red. See this map of
the Milky Way that compares visible light to infrared: http://www.mpi-hd.mpg.de/hfm/HESS/
public/PressRelease/GalScanPress/plots/HESS_Optical_Infrared.gif
5. We know the sun puts out about L = 4 × 1026 W of power. The radius of the sun is about
R = 7 × 108 m. We showed in class that for L = 4πr2 σT 4 , the sun’s surface temperature must
be around 6000 K (you should check that).
a) The sunlight travels outward in a sphere. At the distance to the Earth (R = 1.5 × 1011 m),
how much power passes through each square meter of area?
The area of that large sphere is A = 4πR2 = 2.83 × 1023 m2 . Divide L by this number to
get about 1400 W/m2 .
b) The radius of the Earth is R⊕ = 6.4 × 106 m. It presents a circle toward the sun with an
2
area of πR⊕
. How much total energy hits the Earth each second?
The cross-sectional area of the Earth (not cross-section!) is therefore 1.29 × 1014 m2 . If we
multiply this times our answer from part (a) we get 1.8 × 1017 W.
c) If the Earth can be approximated as a blackbody, then all this power must be reradiated
2
according to L = 4πR⊕
σT 4 . If it were to radiate less, it would warm up, and if it were to radiate
more, it would cool down. What T would the Earth have, to be able to radiate exactly the power
given in part (b)? Is this temperature reasonable?
We solve for T to get
T =
L
2
4πR⊕
σ
!1/4
=
1.8 × 1014
4π × (6.4 × 106 )2 × 5.67 × 10−8
!1/4
K,
(6)
which works out to T = 280 K. That is just a bit above the freezing point of water, which is not
an unreasonable number for the average temperature of the Earth!
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