Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 half-life 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 half-life 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 half-life 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 half-life 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 half-life 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

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.

Ask a homework question - tutors are online