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...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details