Problem#1
An Oceanographic Tracer. Nuclear weapons tests in the 1950s and 1960s released significant amounts of radioactive tritium (3H1, half-life 12.3 years) into the atmosphere. The tritium atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, 2He3 to the
remaining tritium in the water. For example, if the ratio of 2He3 to 1H3 in a sample of water is 1:1, the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of 2He3 to 1H3 is 4.3 to 1.0. How many years ago did this water sink below the surface?
Answer:
The tritium (H-3) decays to He-3. The ratio of the number of He-3 atoms to H-3 atoms
allows us to calculate the time since the decay began, which is when the H-3 was formed by the nuclear
explosion. The H-3 decay is exponential.
The number of tritium (H-3) nuclei decreases exponentially as NH = N0,He-λt, with a half-life
of 12.3 years. The amount of He-3 present after a time t is equal to the original amount of tritium minus the number of tritium nuclei that are still undecayed after time t.
The number of He-3 nuclei after time t is
NHe = N0,H – NH = N0,H – N0,He-λt = N0,H(1 – e-λt)
Taking the ratio of the number of He-3 atoms to the number of tritium (H-3) atoms gives
NHe/NH = [N0,H(1 – e-λt)]/(N0,He-λt) = e-λt – 1
Solving for t gives
t = {ln(1 + NHe/NH)}/λ.
Using the given numbers and T1/2 = ln2/λ, we have
λ = ln2/T1/2 = ln2/12.3 y = 0.0563/y
and t = {ln(1 + 4.3)}/(0.0563/y) = 30 years
Problem#2
Consider the fusion reaction 1H2 + 1H2 → 2He3 + 0n
Answer:
(a) The radius of 2H1 is R = (1.2 x 10-15 m)(2)1/3 = 1.51 x 10-15 m. The barrier energy is the Coulomb potential energy of two 1H2 nuclei with their centers separated by twice this distance:
U = ke2/r = (8.988 x 109 N.m2/C2)(1.602 x 10-19 C)/(2 x 1.51 x 10-15 m) = 7.64 x 10-14 J = 0.48 J
(b) Find the energy equivalent of the mass decrease.
2H1 + 2H1 → 3He2 + 0n1
If we use neutral atom masses there are two electrons on each side of the reaction equation, so their masses cancel. The neutral atom masses are given in Table 43.2.
2H1 + 2H1 has mass 2(2.014102u) = 4.028204u
3He2 + 0n1 has mass 3.016029u + 1.008665u = 4.024694u
The mass decrease is
4.028204 u – 4.024694 u = 3.510 x 10-3u
This corresponds to a liberated energy of
(3.510 x 10-3u)(931.5 MeV/u) = 3.270 MeV, or
(3.270 x 106 eV)(1.602 x 10-19 J/eV) = 5.239 x 10-13 J.
Problem#3
In the 1986 disaster at the Chernobyl reactor in the Soviet Union (now Ukraine), about 1/8 of the 137Cs present in the reactor was released. The isotope 137Cs has a half-life for β decay of 30.07 y and decays with the emission of a total of 1.17 MeV of energy per decay. Of this, 0.51 MeV goes to the emitted electron and the remaining 0.66 MeV to a γ ray. The radioactive 137Cs is absorbed by plants, which are eaten by livestock and humans. How many 137Cs atoms would need to be present in each kilogram of body tissue if an equivalent dose for one week is 3.5 Sv? Assume that all of the energy from the decay is deposited in that 1.0 kg of tissue and that the RBE of the electrons is 1.5.
Answer:
In terms of the number ΔN of cesium atoms that decay in one week and the mass m = 1.0 kg, the equivalent dose is
3.5Sv = (∆N/m)[(RBE)γEγ + (RBE)eEe]
1 day = 8.64 x 104 s. 1 year = 3.156 x 107 s.
3.5 Sv = (∆N/m)[(1)(0.66 MeV) + (1.5)(0.51 MeV)] = (2.283 x 10-13 J)(∆N/m), so
∆N = (1.0 kg)(3.5 Sv)/(2.283 x 10-13 J) = 1.535 x 1013.
λ = ln2/T1/2 = 0.693/[(30.07 y)(3.156 x 107 sec/y)] = 7.30 x 10-10 sec-1.
∆N = |dN/dt|t = λNt
So, N = ∆N/λt = (1.535 x 1013)/[(7.30 x 10-10 s-1)(7 days)(8.64 x 104 s/days)] = 3.48 x 1016.
Problem#4
(a) Prove that when a particle with mass m and kinetic energy K collides with a stationary particle with mass M, the total kinetic energy Kcm in the center-of-mass coordinate system (the energy available to cause reactions) is
Kcm = MK/(M + m)
Assume that the kinetic energies of the particles and nuclei are much lower than their rest energies. (b) If Kth is the minimum, or threshold, kinetic energy to cause an endoergic reaction to occur in the situation of part (a), show that
Kth = –Q(M + m)/M
Answer:
The speed of the center of mass is vcm = vm/(m + M) where v is the speed of the colliding
particle in the lab system. Let Kcm ≡ K′ be the kinetic energy in the center-of-mass system. K′ is
calculated from the speed of each particle relative to the center of mass.
Let vm′ and v′M be the speeds of the two particles in the center-of-mass system. Q is the reaction
energy, as defined in Eq. (43.23). For an endoergic reaction, Q is negative.
(a) v’m = v – vm/(m + M) = Mv/(M + m)
v'M = vm/(M + m)
K’ = ½mvm’2 + ½ MvM’2
K’ = ½ (mM2v2)/(M + m)2 + ½ mM2v2/(M + m)2
K’ = {M/2(m + M)}{(mM/m + M) + m2/(M + m)}v2
K’ = [M/(M + m)](½mv2)
K’ = MK/(M + m) ≡ Kcm.
(b) For an endoergic reaction Kcm = –Q(Q< 0) at threshold. Putting this into part (a) gives
–Q = MKth/(M + m)
Kth = –Q(M + m)/M
Problem#5
Calculate the energy released in the fission reaction 235U92 + 0n1 → 54Xe14
Answer:
Calculate the energy equivalent of the mass decrease
∆m = M(235U92) – M(140Xe54) – M(38Sr94) – mn.
∆m = 238.043923u – 139.921636u – 93.915360u – 1.008665u = 0.1983u
E = ∆mc2 = (0.1983)(931.5MeV/u) = 185 MeV
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