PH300
HW10 answer key
1) T&R ch12 #2 What are the number of neutrons and protons for the
following nuclei: 6Li, 15C, 40K, and 64Cu?
6
Li, Z = 3: N = 6 – 3 = 3,
15
C, Z = 6: N = 15 – 6 = 9
40
K, Z = 19: N = 40 – 19 = 21, 64Cu, Z = 29: N = 64 – 29 = 35
2) T&R ch12 #14 Consider two protons in the 27Al nucleus located 2.0
fm apart. How strong does the nuclear force need to be to overcome
the Coulomb force?
The Coulomb force is described as Fc = kq1q2/r2
Since both charged particles are protons q1 = q2 = 1.602x10-19 C
And k = 8.987x109 Nm2/c2 and r = 2.0x10-15 m
Fc = (8.987x109 Nm2/c2)*(1.602x10-19 C)*(1.602x10-19 C)/( 2.0x10-15 m)2 = 57.7 N
3) T&R ch12 #17 What is the energy released when three alpha
particles combine to form 12C?
B(3α→12C) = [3*M(4He) – M(12C)],
where M(4He) = 4.002603 u and M(12C) = 12 u
So B = 0.007809 u = 7.274 MeV
4) T&R ch12 # 27 Show that the mean (or average) lifetime of a
radioactive sample is τ=1/λ=t1/2/ln(2).
dN/dt = -λN → − ∫
∫
dN
= λ dt →
dt
ln(N) = -λt + constant
N(t ) = e − λt + constant = N 0e − λt
Mean lifetime is time τ such that N0 is reduced by 1/e:
N0/e = N0e-λτ
ln(1/e) = -λτ → τ = (-1)/(-λ) = 1/λ it checks out.
Half life is time t1/2 such that N0 is reduced by ½:
N0
= N0e
2
− t1/2
τ
ln(1/2) = -t1/2/τ → τ = (-t1/2)/(-ln 2) = (t1/2)/(ln 2)
5) T&R ch12 #31 Tritium (half-life=12.33 y) is mostly produced for military
purposes. If we (the US) had stopped producing tritium in 1960 with a
2000-kg stockpile, how much would be left in the year 2010?
N0 = 2000 kg, t0 = 0, tf = 2010 – 1960 = 50 years, t1/2 = 12.33 years
N(50 ) =
− ln (τ )⋅50
2000e 12.33
N(50) = 120.312 kg
This answer makes sense because 50 years is approximately 4 half lives,
2000*(1/2)4 = 125
6) T&R ch12 #35 Show directly using masses that protons do not
undergo any of the beta decays.
Q = [M(parent) – M(daughter)]*c2
β-: Must have neutron parent
β+: p+ → n0 + e+ + ν
Q = [1.00728u – 1.00866u – 2*(5.4858)x10-4u]*c2
= -0.002477uc2
Capture: p+ + e+ → n0 + ν
Q = [1.00728u + 5.4858u - 1.00866u]*c2
= -0.000825uc2
Q < 0 in both cases, proton is stable to beta decay.
7) T&R ch 13#31 Calculate the energy released in kilowatt hours from
the fission of 1kg of 235U. Compare this with the energy released
from the combustion of 1kg of coal. The heat of combustion of coal is
about 29,000 BTU/kg.
Assuming 180 MeV/fission (recoverable)
180 ⋅ 10 6 eV 6.022 ⋅ 10 23 atoms 1 mol 1000 g 1.6022 ⋅ 1019 J
×
×
×
×
= 7.39 ⋅ 1013 J or ~ 1014 J from book
1 atom
1 mol
235 g
1 kg
1 eV
2.053x107 kW*hr
What about coal?
1 kg coal ×
29000 BTU 1.06 ⋅ 10 3 J
×
= 3.07 ⋅ 10 7 J
1 kg
1 BTU
8. 54 kW*hr
The two fuel sources are not comparable.
8) Using the link http://www.new.ans.org/pi/resources/dosechart/
estimate your yearly radiation dose. Do not include medical doses.
Let’s say a person lives on a nuclear submarine.
~50 mrem background (Lots of H20) shielding)
~10-20 mrem reactor proximity (Non-Reactor Room Duty)
70 mrem < 33% of the dose received by someone living in a rain forest
Note: This is extremely conservative. In fact, if one were to remain aboard a sub
for the life of the reactor (10 to 15 years) the received dose would be less than a
lifetime of exposure to cosmic & geological background at sea level.
Extra Credit
Extra Credit
T&R ch12 #61 Radon gas in the form of 222Rn is a health hazard
because it is a gas that occurs as a result of one of the naturally
occurring radioactive decay chains. It tends to collect in basements
and can be inhaled by humans.
a) Which decay chain produces this isotope of radon?
b) Show that 222Rn produces five more disintegrations before a
stable isotope is reached.
c) Choose one of the paths of the decay chain from the 222Rn to the
stable isotope and sum the half-lives. Approximate the number of
days it would take for more than half these decays to occur for a
given amount of radon.
a) Simplified decay chain:
238
U decays to 2 α and 230Th
230
Th decays to an α and 226Ra
226
Ra decays to an α and 222Rn
b) Can show stability/instability for β, α decays at each isotope using
binding energies (i.e. Q<0 or Q>0).
Step 1. 222Rn decays to an α and 218Po
Step 2. 218Po decays to an α and 214Pb
Step 3. 214Pb decays via beta emission to 214Bi
Step 4. 214Bi decays to an α and 210Tl
Step 5. 210Tl decays via beta emission to 210Pb
210
Pb has a half life of 22 years
c) Step 1 takes 3.8 days, Step 2 takes 3.1 minutes, Step 3 takes 27
minutes, Step 4 takes 20 minutes and Step 5 takes 1.3 minutes
So the total half life is about 4 days, w/ longest half life at Rn, thus
Rn’s half-life controls the “feed” to shorter lived isotopes and
therefore the production rate of 210Pb.