Problem#1
Calculate the mass defect for the β+ decay of 7N13. Is this decay energetically possible? Why or why not? The atomic mass of 7N13 is 13.005739 u.
Answer:
The decay is energetically possible if the total mass decreases. Determine the nucleus produced by the decay by balancing A and Z on both sides of the equation. 7N13 → +1e0 + 6C13. To avoid confusion in including the correct number of electrons with neutral atom masses, use nuclear masses, obtained by subtracting the mass of the atomic electrons from the neutral atom masses.
The nuclear mass for 7N13 is Mnuc (7N13) = 13.005739u – 7(0.00054858u) = 13.001899u.
The nuclear mass for 6C13 is Mnuc(6C13) = 13.003355u – 6(0.00054858u) = 13.000064u
The mass defect for the reaction is
∆M = Mnuc(7N13) – Mnuc(6C13) – M(+1e0)
∆M = 13.001899u – 13.000064u – 0.00054858u = 0.001286u
Problem#2
The results of activity measurements on a radioactive sample are given in the table. (a) Find the half-life. (b) How many radioactive nuclei were present in the sample at t = 0. (c) How many were present after 7.0 h?
Answer:
Apply |dN/dt| = λN0e-λt, with λ = ln2/T1/2
ln |dN/dt| = lnλN0 – λt
(a) A least-squares fit to log of the activity vs. time gives a slope of magnitude λ = 0.5995 h-1, for a half-life of ln2/λ = 1.16h.
(b) The initial activity is N0λ, and this gives
N0 = (2.00 x 104Bq)/[(0.5995hr-1)(1 hr/3600s)] = 1.20 x 108.
(c) N = N0e-λt = 1.81 x 106.
Problem#3
A person ingests an amount of a radioactive source with a very long lifetime and activity 0.65µCi. The radioactive material lodges in the lungs, where all of the 4.0-MeV α particles emitted are absorbed within a 0.50-kg mass of tissue. Calculate the absorbed dose and the equivalent dose for one year.
Answer:
Assume the activity is constant during the year and use the given value of the activity to find
the number of decays that occur in one year. Absorbed dose is the energy absorbed per mass of tissue. Equivalent dose is RBE times absorbed dose.
For α particles, RBE 20.
(0.63 x 10-6 Ci)(3.7 x 1010 Bq/Ci)(3.156 x 107 s) = 7.357 x 1011 α particles. The absorbed dose is
(7.357 x 1011)(4.0 x 106 eV)(1.602 x 10-19 J/eV)/(0.50 kg) = 0.943 Gy = 94.3 rad.
The equivalent dose is (20)(94.3 rad) = 1900 rem.
Problem#4
Measuring Very Long Half-Lives. Some radioisotopes such as samarium (149Sm) and gadolinium (152Gd) have half-lives that are much longer than the age of the universe, so we can’t measure their half-lives by watching their decay rate decrease. Luckily, there is another way of calculating the half-life, using
Eq. (43.16). Suppose a 12.0-g sample of 149Sm is observed to decay at a rate of 2.65 Bq. Calculate the half-life of the sample in years. (Hint: How many nuclei are there in the 12.0-g sample?)
Answer:
T1/2 = ln 2/λ. The mass of a single nucleus is 149mp = 2.49 x 10-25 kg.
dN/dt = –λN.
N = (12.0 x 10-3 kg)/(2.49 x 10-25 kg) = 4.82 x 1022
dN/dt = –2.65 decays/s
λ = –(dN/dt)/N = (–2.65 decays/s)/(4.82 x 1022) = 5.50 x 10-23/s
T1/2 = ln2/λ = ln2/(5.50 x 10-23/s) = 1.26 x 1022 s
T1/2 = 3.99 x 1014 y
Problem#5
We Are Stardust. In 1952 spectral lines of the element technetium-99 (99Tc) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of 99Tc is 200,000 years. (a) For how many half-lives has the 99Tc been in the red-giant star if its age is 10 billion years? (b) What fraction of the original 99Tc would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and
other nuclear processes. If the 99Tc had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This
knowledge is what led the late astronomer Carl Sagan to proclaim that “we are stardust.”
Answer:
One-half of the sample decays in a time of T1/2.
(a) (10 x 109y)/(200,000y) = 5.0 x 104
(b) (0.5)50,000. This exponent is too large for most hand-held calculators. But (0.5) = 10-0.301,
So (0.5)50,000 = (10-0.301)50,000 = 10-15,000
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