Homework 1 - Solutions

April 3, 2006

Drew Fustin (dafustin@uchicago.edu)

Physics 237 - Nuclear and Elementary Particle Physics

Problem 1

In a sample of one liter of carbon dioxide at STP an average of 5 disintegrations are observed per minute.

146C → 14

7N + e− + νe

Calculate the atomic fraction of 146C present if the mean life of this nucleus is 8267 years.

Solution

We know that the number of 146C atoms in the sample goes as

N (t) = N0e−λt,

where N0 is the number of atoms at time t = 0 and λ is the inverse of the mean lifetime of 146C, λ =

1/ (8267 years). Between a time t and t + ∆t, there are, on average, ∆N decays, where

∆N = N (t)−N (t + ∆t)= N0e

−λt(1− e−λ∆t

).

Now, if ∆t� λ−1, as it is in our case since ∆t = 1 minute, this becomes (after expanding the second exponentto second order)

∆N = λN0e−λt∆t,

giving a rate of disintegrations as

dN

dt= λN0e

−λt. (1)

Let's say that we are currently at a time t = 0, so

dN

dt= λN0

5 s−1 =1

8267 yr·N0,

giving N0 = 2.17× 1010 , which is the number of 146C atoms present in the sample currently. For any counting

interval that is very small compared to the mean lifetime, it can always be approximated that the activity ofthe sample is

dN

dt= λN,

where N is the current number of atoms. We could have just originally used that fact to solve this problem,but instead we just derived that relationship here.

There is one mole, that is 6.022 × 1023 molecules, of a gas in 22.4 liters at STP. Therefore, in one liter ofcarbon dioxide, there are 2.688 × 1022 molecules. This implies that there are 2.688 × 1022 atoms of 12

6C and5.377× 1022 atoms of 16

8O in our sample. Therefore, the atomic fraction of 146C is just the ratio of the amount

of 146C to the amount of 12

6C, or N(146C

)/N

(126C

)= 8.07× 10−13 .

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Problem 2

A sample of gold is exposed to a neutron beam of constant intensity such that 1010 neutrons per second areabsorbed in the reaction:

19779Au + n → 198

79Au + γ

The nuclide 19879Au undergoes β-decay to 198

80Hg with a mean life of 3.89 days. How many atoms of 19879Au will

be present after 6 days of irradiation? How many atoms of 19880Hg will be present at that time, assuming that

the 19880Hg is una�ected by the neutron beam? What is the equilibrium number of 198

79Au atoms?

Solution

Since 19879Au is being created at a rate of R = 1010atoms/s, and these decay with a decay constant of λ =

1/ (3.89 days), the di�erential change in the number of 19879Au atoms in a time dt is given by

dNAu

dt= R− λNAu. (2)

We assume a solution of the form NAu (t) = A + Be−λt and plug this into Eq. 2, �nding

−Bλe−λt = R− λ(A + Be−λt

),

giving A = R/λ. Also noting that at a time t = 0, NAu should be zero, we �nd that B = −R/λ. Therefore,

NAu (t) =R

λ

(1− e−λt

). (3)

After 6 days of irradiation (t = 6 days), we �nd from Eq. 3 that NAu = 2.64× 1015 .

Now, since we are losing 1010 atoms/s of 19779Au , after 6 days we will have 5.184× 1015 less 197

79Au atoms. Ifwe have 2.64× 1015 atoms of 198

79Au after 6 days, then the di�erence must have converted to 19880Hg. Therefore,

the number of 19880Hg atoms after 6 days is NHg = 2.544× 1015 .

Finally, the equilibrium number of 19879Au atoms is achieved as t→∞ in Eq. 3. That is

NequilAu =

R

λ,

so NequilAu = 3.36× 1015 . This can also be seen by noting that since the amount of 198

79Au increases at a rate

R and decays at a rate λNAu, the the equilibrium is reached when these are equal, i.e. R = λNAu.

Problem 3

Natural uranium found in the Earth's crust contains the isotopes 23592U and 238

92U in the atomic ratio 7.3× 10−3

to 1. Assuming that at the time of formation these two isotopes were produced equally, estimate the time sincethe formation given that the mean lives are 1.03× 109 and 6.49× 109 years respectively.

Solution

The numbers of 23592U and 238

92U are given by

N235 (t) = N0e−λ235t

N238 (t) = N0e−λ238t,

where N0 is the same for both since they were produced equally at the time of formation. Here, λ235 =1/

(1.03× 109 years

)and λ238 = 1/

(6.49× 109 years

). The atomic ratio of these two isotopes is 7.3 × 10−3 to

1, so

N235

N238= 7.3× 10−3

e(λ238−λ235)t = 7.3× 10−3,

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giving

t =log 7.3× 10−3

λ238 − λ235,

or t = 6.023× 109 yr . Radioactive techniques such as this, with suiable corrections, are used to estimate the

age of the Earth. Using a similar technique to this, it is believed that the Earth is around 4.5× 109 years old.

Problem 4

Given that the carbon dioxide in Problem 1 is from the same sample of carbon which was �xed in a biologicalspecimen when the 14

6C/126C ratio was 10−12, calculate the age of the specimen.

Solution

If the 146C/12

6C ratio was 10−12 initially, then

N0

(146C

)= 10−12N0

(126C

).

Since there are 2.688×1022 atoms of 126C in the sample currently, and since 12

6C is stable, there were 2.688×1022

atoms of 126C in the sample initially. Therefore, there were N0 = 2.688×1010 atoms of 14

6C in the sample initially.Now, since there are currently N (t) = 2.17× 1010 atoms of 14

6C in the sample we can use

N (t) = N0e−λt

to �nd

t =log

[N(t)N0

]−λ

,

or t = 1770 yr . Radiocarbon dating was actually invented by the University of Chicago's own Willard Libby

in 1949, earning him the Nobel Prize in 1960. He was a member of the Department of Chemisty here, as wellas the Enrico Fermi Institute.

Problem 5

A beam of neutrons of kinetic energy 0.29 eV and intensity 105 s−1 traverses at normal incidence a foil of 23592U

with thickness 10−1 kg/m2. Any neutron-nucleus collision can have one of three possible results:

1. elastic scattering of neutrons: σe = 2× 10−30 m2,

2. capture of the neutron followed by emission of a γ-ray by the nucleus: σc = 7× 10−27 m2,

3. capture of the neutron followed by splitting of the nucleus into two almost equal parts (�ssion): σf =2× 10−26 m2.

Calculate:(a) the attenuation of the neutron beam by the foil;(b) the number of �ssion reactions occurring per second in the foil caused by the incident beam;(c) the �ux of elastically scattered neutrons at a point 10m from the foil and out of the incident beam, assumingisotropic distribution of the scattered neutrons.

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Solution

First of all, note that the kinetic energy is only useful in this problem in determining the cross sections of thethree possible interactions. Since these are given, the energy should not play an explicit role.(a) We �rst need to �nd the total number of 235

92U atoms per unit area in the foil. The atomic mass of 23592U is

A = 235u where u is an atomic mass unit and has the conversion of 1u = 10−3 kg/NA where NA = 6.022×1023

is Avogadro's number. Since the thickness is given as t = 10−1 kg/m2, we know that the total number of 23592U

atoms is

N235 =t

A

=1000tNA

235 kg

= 2.56× 1023 atomsm2

.

Now, each atom has an a�ective area equal to the sum of the cross sections

σT = σe + σc + σf

= 2.7002× 10−26 m2.

Therefore, each neutron that travels through the foil has a probability of collision equal to the product of thenumber of 235

92U atoms and the total cross section, that is

P = N235σT

= 6.92× 10−3.

Now, if there are 105 neutrons per second (which is the intensity I of the beam) passing through the foil, thenumber of collisions per second is

NT = PI

= 692collisions

s.

Therefore, the attenuation A of the beam is simply the ratio of the number of neutrons left after the foil (thatis, those that did not collide) to the number of neutrons before the foil:

A =NT

I,

so A = 6.92× 10−3 , which is the same as the probability of collision.(b) To �nd the number of �ssions per second, we simply need to use the cross section for �ssion alone indetermining the probability of interaction:

Pf = N235σf

= 5.13× 10−3.

Therefore, the number of �ssions per second is

Nf = PfI,

or Nf = 513 fissions/s .

(c) Finally, we need the number of elastic scattering interactions, which we get from the probability of elasticscattering:

Pe = N235σe

= 5.13× 10−7.

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Therefore, the number of elastic scattering interactions per second is

Ne = PeI

= 5.13× 10−2 scatterss

.

Now, if we're looking for the �ux of these neutrons at a distance of R = 10 m, we need the fractional area ofone square meter of surface area on the sphere to the entire surface area of the sphere:

A1 m2

Asphere=

1 m2

4πR2

= 7.96× 10−4.

Then, the total number of scattered neutrons per second passing through one square meter of surface area ata distance of ten meters is just the product of the number of neutrons scattered per second and the fraction ofthe surface area covered:

Φe =Ne

Asphere,

or Φe = 4.08× 10−5 neutrons/m2/s .

Problem 6

(a) A particle counter registers an average of 0.453 counts per second. What is the probability that it registers2 in any one second?(b) A larger counter registers 1296 in ten minutes so that the best estimate of its rate is 2.16 per second.Estimate the error (standard deviation) on this measured rate and the probability that the measurement islow by more than the error. Assume that the Poisson distribution P (n, m) for large m is approximated by aGaussian distribution with mean m and standard deviation m1/2.

Solution

(a) For a Poisson distribution, the probability of recording n counts (where n is an integer) when the expectednumber is µ is

P (n) =µne−µ

n!.

Therefore, if µ = 0.453 counts per second, the probability of observing n = 2 counts per second is P (2) = 0.065.(b) This time, the sample size is large (there are 1296 counts) so the Poisson distribution can be approximatedas a Gaussian distribution. Now, the Gaussian distribution of the number of counts has an average of 1296and a standard deviation of

√1296 = 36. Therefore, the standard deviation σ in the rate is just the standard

deviation in the counts divided by the total time, that is

σ =√

129610 min

= 0.06 s−1.

Now, for a Gaussian distribution, we know that 68.27% of the area under the curve is contained within onestandard deviation of the mean. Therefore, 31.73% of the area is outside of one standard deviation and half ofthis (15.865%) is the area under just one side of the distribution outside of one standard deviation. Therefore,

the probability that the measurement is low by more than one standard deviation is P (< 1σ) = 0.15865 .

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