COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 27, Problem 13QAP
To determine
That at any given instant a sample of radioactive uranium contains many, many different isotopes of atoms that are not uranium.
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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 u
•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?
•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?
Chapter 27 Solutions
COLLEGE PHYSICS
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- *•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…arrow_forwardB. Carbon-14 is a radioactive isotope of carbon that has a half-life of 5600 years. It is used extensively in dating organic material that is tens of thousands of years old. Model the differential equation of the decay of the isotope if m is the mass and t is the time Calculate the constant using the half-life time. What fraction of the original amount of Carbon-14 in a sample would be present after 10,000 years? i) ii)arrow_forwardA rock contains the radioactive isotope ? − 40 that decays with a half-life of 1.3 ??????? ????? and the radioactive isotope ?? − 87 that decays with a half-life of 48.6 ??????? ?????. The rock is 100 ??????? ????? old. The rock currently has 1000 times as many ?? − 87 atoms as there are ? − 40 atoms. (That is, ??? = 1000 × ??.) When the rock first formed it had ? times as many ?? − 87 atoms as ? − 40 atoms. (That is, ???,0 = ? × ??,0.) What is the value of ??arrow_forward
- • The element copper has naturally occurring isotopes of 63 (62.93amu) and 65 (64.93amu). With an average atomic mass of 63.55amu, calculate the percent abundance of each isotopearrow_forwardStrontium - 90 has a half-life of 25 years.How long will it take 10 grams of Strontium – 90 to decay to 3 grams? Round your answer to the nearest hundredth. time A = A.•() half-life final amount initial amount This is the split factor. After a half-life, one pound becomes pound. Your answer: O not enough information is given O 0.03 years O 43.42 yearsarrow_forwardIn a real or imaginary nucleus of 45X⁹7, (a) how many protons are in the nucleus, (b) how many neutrons are in the nucleus, and (c) how many electrons are in orbit about the nucleus, assuming the atom is electrically neutral? (a) Number (b) Number i (c) Number Units Units Unitsarrow_forward
- -Write an expression for the conservation of momentum in the y direction, taking upwards as positive. -Write the correct equation from below for θTh. -Find the numerical value of θTh in degrees. You may assume that the mass of these particles is the number of nucleons (4 or 238) times the mass of a proton, 1.673 × 10-27 kg. -Write an expression for the speed of the thorium nucleus in terms of sin(θTh).arrow_forwardPart A 90 You are examining a sample of rock that contains 1.6x1010 200 Th atoms decay by alpha emission with a half-life of 77,000 yrs. What is the daughter element produced by this? 226 Ra 88 226 Rn 86 ▾ Part B 230 Rn 86 Cannot answer without a periodic table. 228 Ra 88 Submit Previous Answers All attempts used; correct answer displayed By comparing the number of 330 Th atoms to the number of daughter nuclides, you find that 90% of the sample has decayed to produce the daughter element. How old is the sample? 90 ΠΕ ΑΣΦ SHID ? yearsarrow_forward[“] An element X of atomic number 98 undergoes alpha-decays. The constant rate of alpha decay is n particles per second. (Atomic mass number of X is 251 u). If the element Y (daughter nucleus) has atomic mass of 239 u. What is the mass of total number of alpha particles radiated in grams; (2) 49 (b) 12 g () 2x 10723 g d 4x 10723 garrow_forward
- The rate at which radon is being created is the rate at which radium is decaying, namely A1 N1 or A1 Noe-Ait. But the radon is also decaying at the rate A,N2. Hence, we have dN2 = \1N1 – A2N2, dt or dN2 + A2N2 = A1N1 = 11 Noe¬A1t_ dt This equation is of the form (3.1), and we solve it as follows: I = / A2 dt = Azt, (3.10) N2e^2t A, Noe-Aite^2t dt + c = e(A2-A1)t dt + c= d1 No (A2-A1)t + c, A2 - A1 12, see Problem 19.) Since N2 = 0 at t = 0 (we if A1 + d2. (For the case A1 assumed pure Ra at t = 0), we must have d1 No A1 No 0 = d2 – di or c = Substituting this value of c into (3.10) and solving for N2, we get d, No (e- d2 - A1 - e-Aat). Ait N2arrow_forward•30 Verify that the fusion of 1.0 kg of deuterium by the reaction ?H + ?H - 'He + n could keep a 100 W lamp burning for 2.5 x 10' y. (Q = +3.27 MeV)arrow_forwardPart A Some atomic masses Particle Symbol Mass (u) Electron e 0.00055 Part B Proton 1.00728 Neutron n 1.00866 Part C Hydrogen 1.00783 Helium 4 Не 4.00260 Part D Part E Calculate the binding energy EB of the helium nucleus He. Express your answer in megaelectron volts to three significant figures. • View Available Hint(s) EB = MeVarrow_forward
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