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Unformatted text preview: cy as given in the problem statement. 1616 NUCLEAR PHYSICS AND RADIOACTIVITY SOLUTION The age of the fossils is t=– A 0.100 Bq 5730 yr
ln = –
ln =
A 0.693
0.693 0.23 Bq 0
T1/ 2 6900 yr The maximum error can be found as follows:
When there is an error of +10.0 %, A = 0.100 Bq + 0.0100 Bq = 0.110 Bq , and we have t=– 0.110 Bq 5730 yr
ln = 6100 yr
0.693 0.23 Bq Similarly, when there is an error of –10.0 %, A = 0.100 Bq – 0.0100 Bq = 0.090 Bq , and we
have
5730 yr
0.090 Bq
t=–
ln
= 7800 yr
0.693
0.23 Bq F
G
H I
J
K The maximum error in the age of the fossils is 7800 yr – 6900 yr = 900 yr . 52. REASONING The number of radioactive nuclei remaining after a time t is given by
Equations 31.5 and 31.6 as
N = N 0 e −0 . 693 t / T1/2
where N0 is the number of radioactive nuclei present at t = 0 s and T1/2 is the half life for the
decay. The activity A is proportional to the number of radioactive nuclei that a sample
contains, so it follows that
−0 . 693 t / T1/2
(1)
A = A0 e
We know that the activity for the ancient carbon in the sample is 0.011 Bq per gram of
carbon, whereas for the fresh carbon it is 0.23 Bq per gram of carbon. Knowing the
percentage composition of the contaminated sample, we can determine its activity A by
using the given percentages with the known activities:
AContaminated = 0.980 AAncient + 0.020 AFresh (2) The true age and the apparent age can be obtained by applying Equation (1) to the
corresponding activities.
SOLUTION a. Using Equation (1), we find that the true age is Chapter 31 Problems AAncient = A0 e −0 . 693 t / T1/2 or b 1617 g 0.011 Bq = 0.23 Bq e −0 . 693 t / e
5730 yr j Taking the natural logarithm of both sides of this result gives ln F.011 Bq I = − 0.693 t
0
G.23 Bq J 5730 yr
0
H
K
F.011 Bq I
0
−b
5730 yr g G
ln
0
H.23 Bq J
K t= = 25 000 yr 0.693 b. Using Equation (2) to find the activity of the contaminated sample, we obtain AContaminated = 0.980 AAncient + 0.020 AFresh b g b g = 0.980 0.011 Bq + 0.020 0.23 Bq = 0.0154 Bq
Using this activity in Equation (1), we find that the apparent age of the sample is 0
g F.00154Bq I
G .23 Bq J
H
K b − 5730 yr ln
t= = 22 000 yr 0.693 53. REASONING The atomic mass given for 206 Pb includes the 82 electrons in the neutral
82
atom. Therefore, when computing the mass defect, we must account for these electrons. We
do so by using the atomic mass of 1.007 825 u for the hydrogen atom 1 H , which also
1
includes the single electron, instead of the atomic mass of a proton. To obtain the binding
energy in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV.
SOLUTION
a. Noting that the number of neutrons is 206 – 82 = 124, we can obtain the mass defect ∆m
as follows: b gb g ∆ m = 82 1.007 825 u + 124 1.008 665 u − 205.974 440 u = 1.741 670 u
4
3
1442443 144 244 3 14 244
4
4
82 free hydrogen atoms
(protons plus electrons) 124 free neutrons Intact lead atom
(including 82 electrons) b. Since 1 u is equivalent to 931.5 MeV, the binding energy is b Binding energy = 1.741 670 u F5 J
g931.1 uMeV I = 1622 MeV
G
H
K 1618 NUCLEAR PHYSICS AND RADIOACTIVITY c. The binding energy per nucleon is Binding energy per nucleon =
= Binding energy
Number of nucleons
1622 MeV
= 7.87 MeV/nucleon
206 54. REASONING AND SOLUTION Solving Equation 31.2 for A gives
A = r /(1.2 × 10
3 –15 3 If r is doubled, then A will increase by a factor of 2 = 3 m) 8 55. REASONING AND SOLUTION The activity is A = λN. The decay constant is λ= 0.693
=
T1/ 2 0.693 = 4 .17 × 10 −9 s −1 F
5
b.27 yr g3.1561 × 10 s I
G yr J
H
K
7 As discussed in Section 14.1, the number N of nuclei is the number of moles of nuclei times
Avogadro’s number (which is the number of nuclei per mole). Thus,
N= F 0.50 g Ic.02 × 10 mol h= 5.0 × 10
G .9 g / mol J6 4442444
59
1
3
H 4244 K
14
3
23 Number
of moles −1 21 Avogadro's
number Therefore, A = λN = (4.17 × 10–9 s–1)(5.0 × 1021) = 2.1 × 10 13 Bq . 56. REASONING AND SOLUTION
A
a. The decay reaction is: 242 Pu → Z X + 4 He . Therefore, 242 = A + 4, so that A = 238. In
94
2
addition, 94 = Z + 2, so that Z = 92. Thus, the daughter nucleus is
b. The decay reaction is 24
11 238
92 U. A
0
Na → Z X + –1 e . Therefore, 24 = A. In addition, 11 = Z – 1, so that Z = 12. Thus the daughter nucleus is 24
12 Mg . Chapter 31 Problems c. The decay reaction is 13
7 1619 A
N → Z X + 0 e . Therefore, 13 = A. In addition, 7 = Z + 1, so that
1 Z = 6. Thus, the daughter nucleus is 13
6 C 57. REASONING The reaction and the atomic masses are:
211
211
0
e
82 Pb 3 → 14 244
83 Bi + –13
14244
4
4 210.988 735 u 210.987 255 u 11
11
When a 282 Pb nucleus decays into a bismuth 283 Bi nucleus, the number of orbital electrons
in the bismuth atom is the same (82) as that in the parent lead atom; thus, the bismuth atom
is missing one orbital electron. However, the atomic mass for bismuth 211 Bi (210.987 255
83 11
0
u) includes all 83 electrons for the neutral atom. We note that the combination 283 Bi + –1 e
contains 82 + 1 = 83 electrons, so it is this combination that has an atomic mass of
−
210.987 255u. Energ...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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