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31.5). Since the halflife is inversely proportional to the decay constant (see Equation 31.6),
the halflife also remains constant when the mass of the substance increases.
12. N/N0 = 0.63
13. (b) The activity of an isotope depends on the number of nuclei present and the decay
constant of the isotope (see Section 31.6). The decay constant, on the other hand, is
inversely proportional to the halflife (see Equation 31.6). Thus, it is possible for the two
samples to have the same activity, provided the ratio of the number of nuclei to the halflife
is the same for both.
14. (e) Since the halflife is 1 day, onehalf of the sample remains after 1 day, and onehalf of
that amount (or onequarter of the original) remains after 2 days.
15. (a) The activity of a radioactive sample is equal to the number N of nuclei present times the
decay constant λ of the isotope (see Section 31.6). The decay constant is inversely
proportional to the halflife T1/2 (see Equation 31.6). Thus, the activity is proportional to the
ratio N/T1/2, and this ratio is greatest for Sample 1.
16. (d) The halflife is the time required for onehalf of the original nuclei to disintegrate (see
Section 31.6).
17. (b) The 14 C dating technique assumes that the isotope 14 C was ingested by a living
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organism (see Section 31.7). At one time, all of these items were part of a living animal.
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18. (e) The 16 C activity when the animal died was 0.23 Bq per gram of carbon (see Section
31.7). The present activity is 0.10 Bq per gram of carbon, which is less than half of the
original activity. Since the halflife of 14 C is 5730 yr, the bones must be more than 5000
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years old. 19. Age = 6.9 × 103 years Chapter 31 Problems 1585 CHAPTER 31 NUCLEAR PHYSICS AND RADIOACTIVITY
PROBLEMS
1. SSM REASONING For an element whose chemical symbol is X, the symbol for the
A
nucleus is Z X , where A represents the total number of protons and neutrons (the nucleon
number) and Z represents the number of protons in the nucleus (the atomic number). The
number of neutrons N is related to A and Z by Equation 31.1: A = Z + N . SOLUTION For the nucleus 208
82 Pb , we have Z = 82 and A = 208 . a. The net electrical charge of the nucleus is equal to the total number of protons multiplied
by the charge on a single proton. Since the 208 Pb nucleus contains 82 protons, the net
82 electrical charge of the 208
82 Pb nucleus is bg q net = 82 ( +1.60 × 10 –19 C) = +1.31 × 10 –17 C
b. The number of neutrons is N = A – Z = 208 – 82 = 126 .
c. By inspection, the number of nucleons is A = 208 .
d. The approximate radius of the nucleus can be found from Equation 31.2, namely
r = (1.2 × 10 –15 m) A 1/ 3 = (1.2 × 10 –15 m)(208) 1/3 = 7.1 × 10 –15 m
e. The nuclear density is the mass per unit volume of the nucleus. The total mass of the
nucleus can be found by multiplying the mass m nucleon of a single nucleon by the total
number A of nucleons in the nucleus. Treating the nucleus as a sphere of radius r, the
nuclear density is ρ= mtotal
V = mnucleon A
4
3 πr 3 = mnucleon A
4
3 π (1.2 × 10 –15 m) A 1/ 3 3 = mnucleon
4
3 π ( 1.2 × 10 –15 m) 3 Therefore,
1.67 × 10 –27 kg
ρ=4
=
–15
m) 3
3 π ( 1. 2 × 10 2.3 × 10 17 kg / m 3 1586 NUCLEAR PHYSICS AND RADIOACTIVITY 2. REASONING
a. The number of protons in a given nucleus A
ZX is specified by its atomic number Z. A
b. The number N of neutrons in a given nucleus Z X is equal to the nucleon number A (the
number of protons and neutrons) minus the atomic number Z (the number of protons):
N = A − Z (Equation 31.1). c. In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus
is equal to the number of protons in the nucleus.
SOLUTION
a. The number of protons in the uranium b. The number N of neutrons in the 238
92 U nucleus 202
80 Hg c. The number of electrons that orbit the is Z = 92 . nucleus is N = A − Z = 202 − 80 = 122 . 93
41 Nb nucleus in the neutral niobium atom is equal to the number of protons in the nucleus, or 41 . 3. REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
A
X where A represents the number of protons and neutrons (the nucleon number) and Z
Z
represents the number of protons (the atomic number) in the nucleus.
SOLUTION
a. The symbol 195
78 X indicates that the nucleus in question contains Z = 78 protons, and N = A − Z = 195 – 78 = 117 neutrons . From the periodic table, we see that Z = 78 corresponds to platinum, Pt .
b. Similar reasoning indicates that the nucleus in question is sulfur, S , and the nucleus
contains N = A − Z = 32 – 16 = 16 neutrons .
c. Similar reasoning indicates that the nucleus in question is copper, Cu , and the nucleus
contains N = A − Z = 63 – 29 = 34 neutrons .
d. Similar reasoning indicates that the nucleus in question is boron, B , and the nucleus
contains N = A − Z = 11 – 5 = 6 neutrons . Chapter 31 Problems e. Similar reasoning indicates that the nucleus in question is 1587 plutonium, Pu , and the nucleus contains N =...
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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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