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Unformatted text preview: e numbers of the two types of nuclei are equal
( N0A = N0B ) . Taking the natural logarithm of both sides of the equation above shows that ln ( N A / N B ) = – ( λA – λB ) t or λA – λB = – ln ( N A / N B )
t SOLUTION Since N A / N B = 3.00 when t = 3.00 days, it follows that λA – λB = –ln ( 3.00 )
3.00 days = –0.366 days –1 But we need to find the halflife of species B, so we use Equation 31.6, which indicates that
λ = 0.693 / T1 / 2 . With this expression for λ, the result for λA – λB becomes
1
1
0.693 A – B = –0.366 days −1
T 1/ 2 T1/ 2 Since T1B2 = 1.50 days , the result above can be solved to show that T1A2 = 7.23 days .
/
/ 46. REASONING According to Equation 31.5, the number of nuclei remaining after a time t is
N = N 0 e – λt . If we multiply both sides of this equation by the decay constant λ , we have λN = λN 0 e – λt . Recognizing that λN is the activity A, we have A = A0 e – λt , where A0 is
the activity at time t = 0. A0 can be determined from the fact that we know the mass of the
specimen, and that the activity of one gram of carbon in a living organism is 0.23 Bq. The
decay constant λ can be determined from the value of 5730 yr for the halflife of 14 C using
6
Equation 31.6. With known values for A0 and λ , the given activity of 1.6 Bq can be used
to determine the age t of the specimen. Chapter 31 Problems SOLUTION For 14
6 1613 C , the decay constant is λ= 0.693
0.693
=
= 1.21 × 10 –4 yr –1
T1/ 2
5730 yr The activity at time t = 0 is A0 = ( 9 .2 g)(0.23 Bq / g) = 2.1 Bq.
A0 = 2 .1 Bq , the age of the specimen can be determined from
A = 1.6 Bq = (2.1 Bq) e –(1.21 × 10 –4 Since A = 1.6 Bq and yr –1 ) t Taking the natural logarithm of both sides leads to
ln F.6 Bq I = – (1.21 × 10
1
G Bq J
2.1
HK –4 yr –1 ) t Therefore, the age of the specimen is ln
t= 1.6
F Bq I
G Bq J
2.1
HK –1.21 × 10 –4 yr –1 = 2.2 × 10 3 yr 47. SSM REASONING AND SOLUTION The answer can be obtained directly from
Equation 31.5, combined with Equation 31.6: N
= e – λ t = e –( 0 . 693 ) t / T1/2 = e – ( 0 . 693 )(41 000 yr)/(5730 yr) = 0.0070
N0
The percent of atoms remaining is 0.70 % . 38
48. REASONING Initially, the rock contains a number N0 of uranium 292 U atoms. After a
time t passes, only 60.0% of these atoms are still present. Therefore the final number N of
38
uranium 292 U atoms in the rock is given by N = 0.600 N0 (1) In general, the relationship between the initial and final numbers N0, N of atoms in a
radioactive sample after a time t is 1614 NUCLEAR PHYSICS AND RADIOACTIVITY N = N 0 e − λt
where λ = (31.5) ln 2
(Equation 31.6) is the decay constant and T1 2 is the halflife of the
T1 2 radioactive nuclei.
SOLUTION Solving Equation 31.5 for the term e−λt and then taking the natural logarithm
of both sides, we are able to solve for t: e − λt = Substituting λ = N
N0 or N
− λ t = ln N0 or N
ln N0 t=− λ ln 2
(Equation 31.6) and Equation (1) into Equation (2) yields
T1 2 0.600 N 0
N
ln ln N0
N
t=− 0 =− λ ln 2 T 12 9 (
) = − T1 2 ln 0.600 = − ( 4.47 ×10 yr ) ln ( 0.600 )
ln 2
ln 2 = 3.29 × 109 yr 49. REASONING AND SOLUTION
a. According to Equations 31.5 and 31.6, the ratio N / N 0 is given by
N
–0.693
–0.693 t / T1/2
=e
=e
N0 5.00
/ 5730
c yr hc yr h = 0.999 b. Similarly, we find
N
–0.693 c
3600 s h 122.2 s h
/c
–0.693 t / T1/2
=e
=e
= 1.36 × 10 –9
N0 c. Similarly, we find
N
– 0.693 c yr h 12.33 yr h
5.00
/c
–0.693 t / T1/2
=e
=e
=
N0 0.755 (2) Chapter 31 Problems 1615 50. REASONING The halflife T1 2 of a given isotope is directly related to the decay constant λ by T1 2 = 0.693 λ (Equation 31.6). The decay constant is found from N = N 0e−λt (Equation 31.5), where N and N0 are, respectively, the final and initial numbers of nuclei in the sample,
and t is the time interval. We do not know either the initial or the final number of nuclei in
the sample, but we are told that the final number N is 93.8% of the initial number N0.
Therefore, we have that N = 0.938 N0. SOLUTION Solving N = N 0e−λt (Equation 31.5) for the decay constant λ, we obtain N
= e − λt
N0 N
ln = −λ t N0 or Substituting Equation (1) into T1 2 = T1 2 = 0.693 λ 0.693 λ = or λ =− ln ( N N 0 )
t (1) (Equation 31.6), we find that 0.693
0.693 t
=−
ln ( N N 0 ) ln ( N N 0 ) − t (2) Substituting the time t = 4.51×109 yr and N = 0.938 N0 into Equation (2) yields T1 2 = − ( 0.693) ( 4.51 × 109 yr )
0.693 t
=−
= 4.88 × 1010 yr
ln ( 0.938)
ln 0.938 N 0 N 0 ( ) 51. SSM REASONING According to Equation 31.5, N = N 0 e – λ t . If we multiply both sides
by the decay constant λ, we have λ N = λ N 0e – λt or A = A0 e – λ t where A0 is the initial activity and A is the activity after a time t. The decay constant λ is
related to the halflife through Equation 31.6: λ = 0 . 693 / T1/ 2 . We can find the age of the
fossils by solving for the time t. The maximum error can be found by evaluating the limits
of the accura...
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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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