Unformatted text preview: N 0e − λ t (31.5) where N0 is the number of radioactive nuclei present at t = 0 s, and λ is the decay constant.
The decay constant is related to the halflife T1/2 of the nuclei by λ = 0.693/T1/2 (Equation
31.6). Substituting this expression for λ into Equation 31.5 gives
N = N 0e −0 . 693 t / T1/2 (1) SOLUTION Applying Equation (1) to each type of nucleus, we obtain
NSr
N Cs = N 0, Sr e
N 0, Cs e −0.693 t / T1/ 2, Sr
−0.693 t / T1/ 2, Cs ( = 7.80 × 10 −3 ) e
e −0.693 (15.0 yr ) / ( 29.1 yr )
−0.693 (15.0 yr ) / ( 2.06 yr ) = 0.848 1606 NUCLEAR PHYSICS AND RADIOACTIVITY 37. SSM REASONING AND SOLUTION According to Equation 31.5, N = N 0 e – λ t , the
decay constant is λ=– 1
ln
t FN I = – 1 ln F × 10 I = 0.0866 days
8.14
G J 20 days G × 10 J
N
H
K
4.60
HK
14 –1 15 0 The halflife is, from Equation 31.6, T1/ 2 = 0.693 λ = 0.693
= 8.00 days
0.0866 days −1 38. REASONING AND SOLUTION According to Equation 31.5, the fraction of an initial
sample remaining after a time t is N / N0 = e – λ t , where λ is the decay constant. The decay
constant is related to the halflife T1/ 2 . According to Equation 31.6, the decay constant is λ = 0.693 / T1 / 2 . Therefore, the fraction remaining is
N
–0.693( 30.0 days )/ (8.04 days ) = 0.0753
= e –0.693t /T1/2 = e
N0
This fraction corresponds to a percentage of 7.53 % . 39. SSM REASONING We can find the decay constant from Equation 31.5, N = N 0e – λt . If
we multiply both sides by the decay constant λ , we have λ N = λ N0e – λ t or A = A0e – λ t where A0 is the initial activity and A is the activity after a time t. Once the decay constant is
known, we can use the same expression to determine the activity after a total of six days. SOLUTION Solving the expression above for the decay constant λ , we have λ=– A 285 disintegrations/min 1
1
–1
ln = –
ln = 0.167 days
A t
2 days 398 disintegrations/min 0 Then the activity four days after the second day is A = (285 disintegrations/min) e – (0.167 days –1 )(4.00 days) = 146 disintegrations/min Chapter 31 Problems 1607 ∆N
of the lottery on the second day is the number of
∆t
contestants per day eliminated on the second day. To determine the activity, we will find the
number N of contestants that remains after the first day, and multiply that number by
10% per day (= 0.10 d−1), because 10% of the existing contestants are eliminated per day: 40. REASONING The “activity” ∆N (
= 0.10 d −1 ) N
∆t (1) ∆N
∆N
= λN
via
∆t
∆t
(Equation 31.4) where N is the number of nuclei present in a given sample. Comparison of
Equations 31.4 and (1) will allow us to determine the decay constant λ of the lottery. Once
0.693
(Equation 31.6) to
we have determined the decay constant, we will use T1 2 = In nuclear decay, the decay constant λ is related to the activity λ calculate the halflife of the lottery. SOLUTION
a. On the first day, 10% of the original 5800 contestants are eliminated, and 90% advance
to the next round. Therefore, the number N of contestants at the beginning of the second day
is N = 0.90(5800) = 5220. Using Equation (1), the activity on the second day is
∆N (
= 0.10 d −1 ) N = (0.10 d −1 )(5220) = 522 eliminations/d
∆t
∆N
= λ N (Equation 31.4), we see that the factor 0.10 d−1,
∆t
which is each contestant’s chance of being eliminated from the lottery per day, plays the b. Comparing Equation (1) to role of the decay constant. Therefore, we have that λ = 0.10 d −1 . c. From Equation 31.6, the halflife of the lottery is
T1 2 = 0.693 λ = 0.693
= 6.9 d
0.10 d −1 1608 NUCLEAR PHYSICS AND RADIOACTIVITY 41. REASONING 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 by Activity = ∆N
= λN
∆t (31.4) where N is the number of radioactive nuclei present. The minus sign has been removed from
Equation 31.4, since we have taken the absolute value of both sides of this equation. The
decay constant is related to the halflife T1/2 by λ = 0.693/ T1/2 (Equation 31.6). As
discussed in Section 14.1, the number N of radium nuclei is the number n of moles of nuclei
times Avogadro’s number NA (which is the number of nuclei per mole). Thus, the activity
can be expressed as
0.693
(1)
Activity = λ N =
( nNA )
T1/ 2
However, the number of moles is equal to the mass of radium (one gram) divided by its
mass per mole. Recall that the mass (in grams) per mole has the same numerical value as the
atomic mass of the substance. Since the atomic mass of radium is 226 u, its mass per mole is
226 g/mol.
SOLUTION Using Equation (1), we find that the activity of one gram of radium is Activity = = 0.693
( nNA )
T1/ 2
0.693 3.156 × 107
1.6 × 103 yr 1 yr ( ) ( ) 1.00 g 23
−1 6.02 × 10 mol 226 g/mol 144 2444
4
3
s 14 244
4
3 Avogadro's
Number
number of moles = 3.7 ×1010 disintegrations/s 42. REASONING The heat Q needed to melt a mass m of ice is given by Q = mLf (Equation
12.5), where Lf = 33.5 × 104 J/kg is the latent h...
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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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