COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 27, Problem 94QAP
To determine
Is the reaction
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•49 SSM Generally, more massive nuclides tend to be more un-
stable to alpha decay. For example, the most stable isotope of ura-
nium, 28U, has an alpha decay half-life of 4.5 x 10° y. The most stable
isotope of plutonium is 24Pu with an 8.0 x 10' y half-life, and for
curium we have 248Cm and 3.4 x 10 y. When half of an original sam-
ple of 238U has decayed, what fraction of the original sample of (a) plu-
tonium and (b) curium is left?
*•58 Two radioactive materials that alpha decay, 238U and 232Th,
and one that beta decays, "K, are sufficiently abundant in granite
to contribute significantly to the heating of Earth through the de-
cay energy produced. The alpha-decay isotopes give rise to decay
chains that stop when stable lead isotopes are formed. The isotope
4"K has a single beta decay. (Assume this is the only possible decay of
that isotope.) Here is the information:
Stable
Decay
Half-Life
End
Parent Mode
(y)
Point
(MeV) (ppm)
238U
232Th
4.47 x 10°
206рЬ
51.7
1.41 x 1010
208Pb
42.7
13
1.28 x 10°
40Ca
1.31
4
In the table Q is the total energy released in the decay of one par-
ent nucleus to the final stable end point and f is the abundance of
the isotope in kilograms per kilogram of granite; ppm means parts
per million. (a) Show that these materials produce energy as heat
at the rate of 1.0 x 10-9 W for each kilogram of granite. (b)
Assuming that there is 2.7 x 102 kg of granite in a 20-km-thick
spherical shell at…
•22 O An a particle (*He nucleus) is to be taken apart in the fol-
lowing steps. Give the energy (work) required for each step: (a) re-
move a proton, (b) remove a neutron, and (c) separate the remain-
ing proton and neutron. For an a particle, what are (d) the total
binding energy and (e) the binding energy per nucleon? (f) Does
either match an answer to (a), (b), or (c)? Here are some atomic
masses and the neutron mass.
"He 4.002 60 u
2H 2.014 10 u
3H 3.016 05 u
'H 1.007 83 u
1.008 67 u
Chapter 27 Solutions
COLLEGE PHYSICS
Ch. 27 - Prob. 1QAPCh. 27 - Prob. 2QAPCh. 27 - Prob. 3QAPCh. 27 - Prob. 4QAPCh. 27 - Prob. 5QAPCh. 27 - Prob. 6QAPCh. 27 - Prob. 7QAPCh. 27 - Prob. 8QAPCh. 27 - Prob. 9QAPCh. 27 - Prob. 10QAP
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- •3 @ A thermal neutron (with approximately zero kinetic energy) is absorbed by a 23U nucleus. How much energy is transferred from mass energy to the resulting oscillation of the nucleus? Here are some atomic masses and the neutron mass. 237U 237.048 723 u 239U 239.054 287 u 238U 238.050 782 u 240U 240.056 585 u 1.008 664 uarrow_forward•14 A 236U nucleus undergoes fission and breaks into two mid- dle-mass fragments, 140Xe and 96Sr. (a) By what percentage does the surface area of the fission products differ from that of the origi- nal 236U nucleus? (b) By what percentage does the volume change? (c) By what percentage does the electric potential energy change? The electric potential energy of a uniformly charged sphere of ra- dius r and charge Q is given by 3 514περarrow_forward8. The binding energy & d.ee Mev respectively. • of tritium (H) and deuteron (24) released when 340 & Date Page I are what will be energy 24 fuse to form a stable tHe of binding energy 28.3 MeV ? O 2 15 M me AMSarrow_forward
- 6 Carbon-14 (C) dating is a method for finding the age of an organic artifact from the quantity of 14C it contains. Carbon-14, an unstable isotope of carbon, follows a well-known sequence of decay processes. The decay constants of these processes have been well established, allowing researchers to determine the age of an artifact knowing both the original amount of 14C and the current amount. In the lab, it is relatively easy to measure the activity of a sample and to estimate the mass of carbon in the sample. From these measurements, it is possible to find the age of the sample.arrow_forward•61 The isotope 238SU decays to 206 Pb with a half-life of 4.47 x 10° y. Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, 238U → 206Pb + various decay products. A rock is found to contain 4.20 mg of 23$U and 2.135 mg of 206PB. Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) 238U and (b) 206Pb does the rock now contain? (c) How many atoms of 238U did the rock contain at formation? (d) What is the age of the rock?arrow_forwardA nuclear physicist finds 1.0of 236Uin a piece of uranium ore (T1/2=2.348107y) . (a) Use die decay law to determine how much 236Uwould had to have been on Earth when it formed 4.543109yago for 1.0gto be left today, (b) What is unreasonable about this result? (c) How is this unreasonable result resolved?arrow_forward
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