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Unformatted text preview: y is released during β decay, so the combined mass of the bismuth
211
211
−
83 Bi daughter nucleus and the β particle is less than the mass of the lead 82 Pb parent
nucleus. The difference in mass is equivalent to the energy released. To obtain the energy in
MeV, we will use the fact that 1 u is equivalent to 931.5 MeV. SOLUTION The mass difference that accompanies the β − decay is given by 210.988 735 u − 210. 987 255 u = 1.48 × 10
energy released is ( −3 u. Since 1 u is equivalent to 931.5 MeV, the ) 931.5 MeV Energy released = 1.48 ×10−3 u = 1.38 MeV
1u 58. REASONING In the radiocarbon method, the number of radioactive nuclei remaining at a
given instant is related to the number present initially, the time that has passed since the
individual died, and the decay constant for 14 C . Thus, to determine how long ago the
6
individual died, we will need information about the number of nuclei present in the material
discovered with the mummy and the number present initially, which can be related to the
activity of the material found with the body and the initial activity. We will also need the
decay constant, which can be obtained from the halflife of 14 C .
6 1620 NUCLEAR PHYSICS AND RADIOACTIVITY SOLUTION The number N of radioactive nuclei present at a time t is N = N 0 e −λ t (31.5) where N0 is the number present initially at t = 0 s and λ is the decay constant for
Solving this equation for t, we find that
N
= e −λ t
N0 or N
= −λ t
ln N 0 or 14
6C . 1 N t = − ln λ N0 Since the activity A is proportional to the number N of radioactive nuclei, this expression for
t becomes
1 N 1 A (1)
t = − ln = − ln λ N0 λ A0 The decay constant is related to the halflife T1/2 according to λ= 0.693
T1/2 (31.6)
Substituting this expression into Equation (1) reveals that A
1
1 A t = − ln = − ln λ A0 0.693 / T1/2 A 0 Noting that A/A0 = 0.785 and that T1 / 2 = 5730 yr , we find that
T A 5730 yr 3
t = − 1/2 ln = − ln ( 0.785 ) = 2.00 ×10 yr
A 0.693 0.693 0 59. SSM WWW REASONING AND SOLUTION As shown in Figure 31.18, if the first
dynode produces 3 electrons, the second produces 9 electrons ( 3 2 ), the third produces 27
electrons ( 3 3 ), so the Nth produces 3N electrons. The number of electrons that leaves the
14th dynode and strikes the 15th dynode is
314 = 4 782 969 electrons Chapter 31 Problems 1621 60. REASONING The mass m of a substance is equal to the number n of moles times the mass
per mole (see Section 14.1). The mass per mole (in grams per mole) has the same numerical
value as the atomic mass of the substance. Since the atomic mass of gold is 197.968 u, its
mass per mole is 197.968 g/mol. Thus, m = n (197.968 g/mol). Recall also that the number n
of moles is equal to the number N of gold atoms divided by Avogadro’s number NA;
n = N /NA. Therefore, the mass of gold can be written as
N
m = n (197.968 g/mol ) = (197.968 g/mol ) NA (1) The activity ∆N / ∆t of a radioactive sample is the number of disintegrations per second
that occurs. The activity is related to the decay constant λ and the number N of atoms by Activity = ∆N
= λN
∆t (31.4) The minus sign has been removed from Equation 31.4, since we have taken the absolute
value of both sides of this equation. Solving Equation 31.4 for N and substituting the result
into Equation (1), we find ∆N (2)
m = ∆t (197.968 g/mol )
λN A
Finally, we recognize that the decay constant λ is related to the halflife T1/2 by
λ = 0.693/ T1/2 (Equation 31.6). Substituting this expression for λ into Equation (2) gives
∆N
T (197.968 g/mol )
∆t 1/ 2
m=
( 0.693) NA SOLUTION Expressing the activity of radioactive gold in becquerels (1 Ci = 3.70 × 1010
4
Bq) and the halflife in seconds (1 day = 8.64 × 10 s), we find that the mass of gold
required to produce an activity of 315 Ci is 1622 NUCLEAR PHYSICS AND RADIOACTIVITY ∆N
T (197.968 g/mol )
∆t 1/ 2
m=
( 0.693) NA
10
4 ( 315 Ci ) 3.70 × 10 Bq ( 2.69 days ) 8.64 × 10 s (197.968 g/mol )
1 Ci 1 day =
23
−1 )
( 0.693) ( 6.02 × 10 mol = 1.29 × 10−3 g 61. REASONING AND SOLUTION
reaction gives The conservation of linear momentum applied to the mαvα + mΤvΤ = 0 (1) The energy released in the decay is assumed to be kinetic, so
(1/2)mαvα2 + (1/2)mΤvΤ2 = 4.3 MeV (2) Solving Equation (1) for vα, substituting into Equation (2) and rearranging, gives the kinetic
energy of the thorium atom.
1
2 2
mT v T = 4 .3 MeV
4 .3 MeV
=
=
mT
234 .0436 u
1+
1+
4 .0026 u
mα 0.072 MeV The kinetic energy of the α particle is, then, 4.3 MeV – 0.072 MeV = 4.2 MeV . 62. REASONING
a. Each α decay process takes the general form
A
A− 4
4
ZP → { + {
Z −2 D
2 He
{
Parent
nucleus Daughter
nucleus α particle Therefore, each α decay reduces the atomic mass number A by 4 and the atomic number Z
by 2. Writing out the overall decay process for four α decays will allow us to determine the
atomic mass number and atomic number of the “greatgreatgranddaughter” of the thorium
228
90Th nucleus. Chapter 31 Problems 1623 b. The total amount of ene...
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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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