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Unformatted text preview: eat of fusion for water (see Table 12.3).
The energy E released when a nucleus disintegrates is related to the mass decrease ∆m by
2
E = (∆m)c (Equation 28.5). A mass decrease of one atomic mass unit (∆m = 1 u)
corresponds to a released energy of 931.5 MeV. The total energy ETotal released is just the
number of disintegrations times the energy for each one, or ETotal = nE. Chapter 31 Problems In a time period equal to one halflife, the number of radioactive nuclei that decay is 1609 1
2 N0 , where N0 is the number present initially.
SOLUTION According to Equation 12.5, the mass of ice melted is m = Q/Lf. The heat Q is
provided by the total energy from the decay or Q = ETotal = nE, as discussed in the
REASONING. The mass of water melted, then, becomes m = nE/Lf . To use this expression,
we need the energy E that is released by one disintegration.
The disintegration process is
224
88 Ra
224.020 186 u → 220
86 Rn + 4
2 He 220.011 3684 + 4.002 603 3
u
1444u 24444
224.013 971 u As usual, the masses are atomic masses and include the mass of the orbital electrons. This
causes no error here, however, because the same total number of electrons is included on the
left and right sides of the arrow in the process above. The decrease in mass, then, is
224.020 186 u – 224.013 971 = 0.006215 u
Since 931.5 MeV of energy corresponds to 1 u and since 1 eV = 1.60 × 10–19 J, the energy
release by one disintegration is b E = 0.006215 u F
g931.5 × 10
G 1u
H 6 eV IF × 10 J I = 9.26 × 10
J 1.601 eV J
G
K
H
K
−19 −13 J Since the time period being considered is equal to one halflife, the number n of
disintegrations that occurs is one half the number of radioactive nuclei initially present. We
can now calculate the mass of ice melted as follows:
m= nE
=
Lf 1
2 2.69
c × 10 h9.26 × 10
c
21 33.5 × 10 J / kg
4 −13 J h
= 3720 kg 43. REASONING Recall from Section 14.1 that the mass m of a substance is equal to the
number n of moles times the mass per mole. The mass (in grams) per mole has the same
numerical value as the atomic mass of the substance, so the mass per mole for strontium is
89.908 g/mol. Thus, m = n (89.908 g/mol). Also, the number n of moles is equal to the
number N of strontium atoms divided by Avogadro’s number NA; n = N /NA. The mass of
strontium can, therefore, be written as 1610 NUCLEAR PHYSICS AND RADIOACTIVITY N
m = n ( 89.908 g/mol ) = ( 89.908 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 m = ∆t ( 89.908 g/mol )
(2)
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 ( 89.908 g/mol )
∆t 1/ 2
m=
( 0.693) NA (3) SOLUTION Using Equation (3), we find that ∆N
T ( 89.908 g/mol )
∆t 1/ 2
m=
( 0.693) NA ( 6.0 ×105 Bq ) ( 29.1 yr ) 3.156 ×10
= 7 s ( 89.908 g/mol ) 1 yr ( 0.693) ( 6.02 × 1023 mol−1 ) = 1.2 × 10−7 g 44. REASONING The average distance x traveled by the neutrons before 75.0% of them
decay is x = vt, where v is their speed and t is the time for 75.0% of them to decay. The
speed of each neutron is related to its kinetic energy KE (= 5.00 eV) by KE = 1 m v 2
2
(Equation 6.2). Solving this equation for the speed gives Chapter 31 Problems v= 1611 2 ( KE )
m Thus, the distance traveled can be written as 2 ( KE )
t
m x = vt = (1) The time can be obtained by noting that the number N of neutrons remaining after a time t is
given by N = N 0e− λ t (Equation 31.5), where N0 is the original number of neutrons and λ is
the decay constant. The decay constant is related to the halflife T1/2 of a neutron by
λ = 0.693/T1/2 (Equation 31.6), so Equation 31.5 can be written as N = N0 0.693 −
t T1 / 2 e Dividing both sides of this equation by N0, taking the natural logarithm, and solving for t
gives
N
=e
N0 0.693 −
t T1 / 2 or N 0.693 ln = − T
t N0 1/ 2 or t=− T1/ 2 N ln 0.693 N 0 (2) SOLUTION Substituting the result for t in Equations (2) into Equation (1) and noting that
N/N0 = 0.750 (since the number of neutrons decreases to 75.0% of its initial value), we
obtain x= 2 ( KE )
2 ( KE ) −T1/ 2 N t= ln m
m 0.693 N 0 1.60 ×10−19 J 60 s 2 ( 5.00 eV ) − (10.4 min ) 1 eV 1 min ln ( 0.750 ) = 8.01× 106 m
= 0.693 1.675 ×10−27 kg 1612 NUCLEAR PHYSICS AND RADIOACTIVITY 45. SSM REASONING According to Equation 31.5, the number of nuclei remaining after a
time t is N = N 0e – λ t . Using this expression, we find the ratio N A / N B as follows:
–λ t NeA
NA
– ( λ – λ )t
= 0A – λ t = e A B
NB
N 0Be B
where we have used the fact that initially th...
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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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