ch40.pdf

CHAPTER 40

Nuclear Physics 1* Give the symbols for two other isotopes of (a) 14N, (b) 56Fe, and (c) 118Sn (a) 15N, 16N; (b) 54Fe, 55Fe; (c) 114Sn, 116Sn 2 Calculate the binding energy and the binding energy per nucleon from the masses given in Table 40-1 for (a) 12C, (b) 56Fe, and (c) 238U. (a) Use Equ. 40-3 and Table 40-1. (b), (c) Proceed as in part (a)

(a) Eb = (61.007825 + 6 1.008665 - 12.00)931.5 MeV = 92.16 MeV; Eb/A = 7.68 MeV (b) Z = 26, N = 30; Eb = 488.1 MeV; Eb/A = 8.716 MeV (c) Z = 92, N = 146; Eb = 1804 MeV; Eb/A = 7.58 MeV

3 Repeat Problem 2 for (a) 6Li, (b) 39K, and (c) 208Pb. (a), (b), (c) Proceed as in Problem 40-2. (a) Z = 3, N = 3; Eb = 31.99 MeV; Eb/A = 5.33 MeV

(b) Z = 19, N = 20; Eb = 333.7 MeV; Eb/A = 8.556 MeV (c) Z = 82, N = 126; Eb = 1636.5 MeV; Eb/A = 7.868 MeV

4 Use Equation 40-1 to compute the radii of the following nuclei: (a) 16O, (b) 56Fe, and (c) 197Au. (a), (b), (c) Use Equ. 40-1 (a) R16 = 3.78 fm; (b) R56 = 5.74 fm; (c) R197 = 8.73 fm 5* (a) Given that the mass of a nucleus of mass number A is approximately m = CA, where C is a constant, find an expression for the nuclear density in terms of C and the constant R0 in Equation 40-1. (b) Compute the value of this nuclear density in grams per cubic centimeter using the fact that C has the approximate value of 1 g per Avogadro's number of nucleons. (a) From Equ. 40-1, R = R0A1/3, the nuclear volume is V = (4p/3)R03A. With m = CA, r = m/V = 3C/4pR03. (b) Given that C = 1/6.021023 g and R0 = 1.510-13 cm, r = 1.181014 g/cm3. 6 Derive Equation 40-2; that is, show that the rest energy of one unified mass unit is 931.5 MeV. 1 u = 1.66054010-27 kg (see p. EP-4). Hence, uc2 = [(2.997924108)21.66054010-27/1.60217710-19] eV = 9.3149108 eV = 931.49 MeV. 7 Use Equation 40-1 for the radius of a spherical nucleus and the approximation that the mass of a nucleus of mass

number A is A u to calculate the density of nuclear matter in grams per cubic centimeter.

Chapter 40 Nuclear Physics

The density of a sphere is r = M/V. In this case M = 1.6610-27A kg and V = (4p/3)(1.510-15)3A m3. Thus r = 1.1741017 kg/m3 = 1.1741014 g/cm3. 8 The electrostatic potential energy of two charges q1 and q2 separated by a distance r is U = kq1q2/r, where k is

the Coulomb constant. (a) Use Equation 40-1 to calculate the radii of 2H and 3H. (b) Find the electrostatic potential energy when these two nuclei are just touching, that is, when their centers are separated by the sum of their radii.

(a) Use Equ. 40-1 (b) Evaluate U; r = 4.05 fm

R2 = 1.89 fm; R3 = 2.16 fm

U = (1.44 eV.nm)/(4.0510-6 nm) = 0.356 MeV

9* (a) Calculate the radii of 56

141

0Ba and 36

920Kr from Equation 40-1. (b) Assume that after the fission of 235U into

141Ba and 92Kr, the two nuclei are momentarily separated by a distance r equal to the sum of the radii found in (a), and calculate the electrostatic potential energy for these two nuclei at this separation. (See Problem 8.) Compare your result with the measured fission energy of 175 MeV. (a) Use Equ. 40-1, with A = 141 and 92, respectively

(b) Use ke2 = 1.44 MeV.fm; U = kZ1Z2e2/(r1 + r2)

For fm 7.81 = ,Ba14156 R 0; for

fm 6.77 = ,Kr9236 R 0

U = (1.445636/14.58) MeV = 199 MeV; this is somewhat greater than the fission energy of 175 MeV

10 Why is the decay series A = 4n + 1 not found in nature? The parent of that series, 237Np, has a half-life of 2 106 y which is much shorter than the age of the earth. There is no naturally occurring Np remaining on earth. 11 A decay by a emission is often followed by b decay. When this occurs, it is by b and not b + decay. Why? Generally, a-decay leaves the daughter nucleus neutron rich, i.e., above the line of stability. The daughter nucleus therefore tends to decay via b- emission which converts a nuclear neutron to a proton. 12 The half-life of 14C is much less than the age of the universe, yet 14C is found in nature. Why? 14C is found on earth because it is constantly being formed by cosmic rays in the upper atmosphere in the reaction 14N + n 14C + 1H. 13* What effect would a long-term variation in cosmic-ray activity have on the accuracy of 14C dating? It would make the dating unreliable because the current concentration of 14C is not equal to that at some earlier 14 Homer enters the visitors chambers, and his geiger-beeper goes off. He shuts off the beep, removes the device

from his shoulder patch and holds it near the only new object in the room: an orb which is to be presented as a gift from the visiting Cartesians. Pushing a button marked monitor, Homer reads that the orb is a radioactive source with a counting rate of 4000 counts/s. After 10 min, the counting rate has dropped to 1000 counts/s. The sources half-life appears on the geiger-beeper display. (a) What is the half-life? b) What will the counting rate be 20 min after the monitoring device was switched on?

(a) R drops by factor of 22 in 10 min. Use Equ. 40-12 (b) Use Equ. 40-12

t1/2 = 5 min R20 = 250 Bq

15 A certain source gives 2000 counts/s at time t = 0. Its half-life is 2 min. (a) What is the counting rate after


4 min? (b) After 6 min? (c) After 8 min? (a), (b), (c) Use Equ. 40-12 (a) R4 = 500 Bq; (b) R6 = 250 Bq; (c) R8 = 125 Bq 16 The counting rate from a radioactive source is 8000 counts/s at time t = 0, and 10 min later the rate is 1000

counts/s. (a) What is the half-life? (b) What is the decay constant? (c) What is the counting rate after 20 min? (a) Use Equ. 40-12 (b) Use Equ. 40-11 (c) Use Equ. 40-12, 20 min = 6t1/2

10 min = 3t1/2; t1/2 = 3.333 min = 200 s

l = 0.693/200 = 3.46510-3 s-1 R = 125 Bq

17* The half-life of radium is 1620 y. Calculate the number of disintegrations per second of 1 g of radium, and show that the disintegration rate is approximately 1 Ci. 1. Use Equ. 40-11 to find l 2. Determine N0 = NA/M 3. Find R from Equ. 40-7

l = (0.693/1620) y-1 = 4.2810-4 y-1 = 1.3510-11 s-1 N0 = 6.021023/226 = 2.6641021

R = (1.3510-11 2.6641021) s-1 = 3.601010 s-1 3.71010 s-1 = 1 Ci

18 A radioactive silver foil (t1/2 = 2.4 min) is placed near a Geiger counter and 1000 counts/s are observed at time t = 0. (a) What is the counting rate at t = 2.4 min and at t = 4.8 min? (b) If the counting efficiency is 20%, how many radioactive nuclei are there at time t = 0? At time t = 2.4 min? (c) At what time will the counting rate be about 30 counts/s? (a) Use Equ. 40-12 (b) N = R/h (c) Use Equ. 40-12; n ln(2) = ln(R0/R)

R2.4 = 500 Bq; R4.8 = 125 Bq N0 = 5000; N2.4 = 2500 n = 5.059; t = 12.14 min

19 Use Table 40-1 to calculate the energy in MeV for the a decay of (a) 226Ra and (b) 242Pu. (a) 1. Write the reaction 2. Find DE from Table 40-1 (b) 1. Write the reaction 2. Find DE from Table 40-1

226Ra 222Rn + 4He

DE = 931.5(226.025360 - 222.017531 - 4.002603) MeV = 4.868 MeV 242Pu 238U + 4He

DE = 6.999 MeV 20 Suppose that two billion years ago 10% of the mass of the earth were 14C. Approximately what percent of the

mass of the earth today would be 14C, neglecting formation of 14C in the atmosphere? 1. Give t in terms of t1/2 2. M(t) = M0(1/2)n

t = 2109/5730 = 3.45105t1/2 = nt1/2 M(t) = 0%

21* At the scene of the crime, in the museums west wing, Angela found some wood chips, so she slipped them into her purse for future analysis. They were allegedly from an old wooden mask, which the guard said he threw at the would-be thief. Later, in the lab, she determined the age of the chips, using a sample which contained 10 g of carbon and showed a 14C decay rate of 100 counts/min. How old are they?


1. Find R/R0 =(1/2)n 2. Solve for n = [ln (R/R0)]/[ln (0.5)] 3. Find t = nt1/2

R0 = (1510 ) s-1 = 150 s-1; (1/2)n = 2/3 n = 0.585 t = (0.5855730) y = 3352 y

22 The thief in problem 21 had been after a valuable carving made from a 10,000 year old bone. The guard said that

he chased the thief away, but Angela suspects that the guard is an accomplice, and that the bone in the display case is in fact a fake. If a sample of the bone containing 15 g of carbon were to be analyzed, what should the decay rate of 14C be if it is a 10,000-year-old bone?

Use Equ. 40-12 to find R; n = 10,000/t1/2 n = 1.745; R = (151560)/21.745 Bq = 4027 Bq 23 Through a friend in security at the museum, Angela got a sample having 175 g of carbon. The decay rate of 14C

was 8.1 Bq. (a) How old is it? (b) Is it from the carving described in problem 22? (a) Use Equ. 40-12 (b) It is probably from the bone of Problem 22

R0/R = (17515)/(8.160) = 5.40; n = [ln(5.40)/ln(2)] = 2.43; T = 13,940 y

24 A sample of a radioactive isotope is found to have an activity of 115.0 Bq immediately after it is pulled from the

reactor that formed it. Its activity 2 h 15 min later is measured to be 85.2 Bq. (a) Calculate the decay constant and the half-life of the sample. (b) How many radioactive nuclei were there in the sample initially?

(a) Use Equ. 40-7 (b) Use Equ. 40-8

l = [ln(R0/R)]/t = 0.133 h-1; t1/2 = 5.20 h N0 = 1153600/0.133 = 3.11106 nuclei

25* Derive the result that the activity of 1 g of natural carbon due to the b decay of 14C is 15 decays/min = 0.25 Bq. 1. Find number of 14C per gram of C 2. Find the decay constant 3. R = lN

N(14C) = (6.021023/12)(1.310-12) = 6.521010

l = (0.693/5730) y-1 = 3.8310-12 s-1 = 2.3010-10 min-1

R = (6.521010 2.3010-10) min-1 = 1.50 min-1


26 Measurements of the activity of a radioactive sample have yielded the following results. Plot the activity as a

function of time, using semilogarithmic paper, and determine the decay constant and half-life of the radioisotope. Time(min) Activity Time(min) Activity 0 4287 20 880 5 2800 30 412 10 1960 40 188 15 1326 60 42

The plot of the data is shown. The data fit a straight line of slope (-2.0/57) min-1 = 0.0351 min-1. The decay constant is therefore l = 0.0351/log(e) min-1 = 0.0808 min-1. The half-life is t1/2 = 8.57 min.

27 (a) Show that if the decay rate is R0 at time t = 0 and R1 at some later time t1, the decay constant is given by l = t1-1ln(R0/R1) and the half-life is given by t1/2 = 0.693t1/ln(R0/R1). (b) Use these results to find the decay constant and the half-life if the decay rate is 1200 Bq at t = 0 and 800 Bq at t1 = 60 s. (a) From Equ. 40-7 it follows that R0/R = elt. Consequently, l = [ln(R0/R)]/t. The half-life is given by Equ. 40-11. Thus, t1/2 = 0.693t/ln(R0/R). (b) Find l and t1/2 for R0/R = 1.5, t = 60 s l = 0.00676 s-1; t1/2 = 102.5 s 28 A wooden casket is thought to be 18,000 years old. How much carbon would have to be recovered from this

object to yield a 14C counting rate of no less than 5 counts/min? 1. Find the decay rate per gram of carbon 2. Determine mass of carbon required

R = 15/2n; n = 18000/5730 = 3.14; R = 1.7 counts/min M = 5/1.7 = 2.94 grams

29* A 1.00-mg sample of substance of atomic mass 59.934 u emits b particles with an activity of 1.131 Ci. Find the decay constant for this substance in s 1 and its half-life in years. 1. Find N 2. l = R/N (see Equ. 40-7)

N = (6.0221023 10-3/59.934) = 1.0051019 R = 1.1313.71010; l = 4.16510-9 s-1


3. Use Equ. 40-11 t1/2 = (0.693/4.16510-9) s = 1.664108 s = 5.27 y 30 The counting rate from a radioactive source is measured every minute. The resulting counts per second are 1000,

820, 673, 552, 453, 371, 305, 250. Plot the counting rate versus time on semilog graph paper, and use your graph to find the half-life of the source. The plot of the data is shown in the figure. From the figure it is apparent that the counting rate drops to 500 Bq at t = 3.5 min.

31 A sample of radioactive material is initially found to have an activity of 115.0 decays/min. After 4 d 5 h, its activity

is measured to be 73.5 decays/min. (a) Calculate the half-life of the material. (b) How long (from the initial time) will it take for the sample to reach an activity level of 10.0 decays/min?

(a) Use the result of Problem 27; 4 d 5 h = 101 h (b) Use Equs. 40-11 and 40-7

t1/2 = [0.693101/ln(115/73.5)] h = 156.4 h = 6 d 12.4 h

l = 0.00443 h-1; t = ln(11.5)/l = 551.2 h = 23 d 32 The rubidium isotope 87Rb is a b emitter with a half-life of 4.9 1010 y that decays into 87Sr. It is used to determine

the age of rocks and fossils. Rocks containing the fossils of early animals contain a ratio of 87Sr to 87Rb of 0.0100. Assuming that there was no 87Sr present when the rocks were formed, calculate the age of these fossils.

1. NSr = N0,Rb - NRb; also NSr = 0.01NRb; find N0,Rb/NRb 2. Determine t = t1/2 ln(1.01)/ln(2)

N0,Rb/NRb = 1.01 t = 7.03108 y

33* If there are N0 radioactive nuclei at time t = 0, the number that decay in some time interval dt at time t is dN =

lN0e lt dt. If we multiply this number by the lifetime t of these nuclei, sum over all the possible lifetimes from t = 0 to t = , and divide by the total number of nuclei, we get the mean lifetime t:

dtetNdtN

tllt -

== 000

1

Show that t = 1/l.


Note that l is a constant; so dxxedtetxt -

-

=00

1l

l l . The definite integral has the value 1, so t = 1/l.

34 Using Table 40-1, find the Q values for the following reactions: (a) 1H + 3H 3He + n + Q and (b) 2H + 2H 3He + n + Q. (a), (b) Use Equ. 40-19 and Table 40-1 (a) Q = (1.007825 + 3.016050 - 3.016030 -

1.008665)(931.5 MeV) = (-0.000820)(931.5 MeV) = -0.7638 MeV (b) Q = (0.003509)(931.5 MeV) = 3.269 MeV

35 Using Table 40-1, find the Q values for the following reactions: (a) 2H + 2H 3H + 1H + Q, (b) 2H + 3He 4He + 1H + Q, and (c) 6Li + n 3H + 4He + Q. (a), (b), (c) Use Equ. 40-19 and Table 40-1 (a) Q = (22.014102 - 3.016050 - 1.007825)(931.5 MeV)

= (0.004329)(931.5 MeV) = 4.032 MeV; (b) Q = 18.35 MeV; (c) Q = 4.785 MeV

36 (a) Use the atomic masses m = 14.00324 u for 6

14C and m = 14.00307 u for

7

14N to calculate the Q value (in MeV)

for the b decay

6

14C

7

14N + b + n e

(b) Explain why you do not need to add the mass of the b to that of atomic 7

14N for this calculation.

(a) Use Table 40-1 Q = (14.00324 - 14.00307)(931.5 MeV) = 0.158 MeV (b) The atomic mass of 14N includes that of 7 electrons, so all electronic masses are taken into account.

37* (a) Use the atomic masses m = 13.00574 u for 7

13N and m = 13.003354 u for

6

13C to calculate the Q value (in

MeV) for the b decay

7

13N

6

13C + b + + ne

(b) Explain why you need to add two electron masses to the mass of 6

13C in the calculation of the Q value for this

reaction. (a) For b + decay, Q = (mi - mf - 2me)c2 Q = (0.002386931.5 - 20.511) MeV = 1.20 MeV

(b) The atomic masses include the masses of the electrons of the neutral atoms. In this reaction the initial atom has 7 electrons, the final atom only has 6 electrons. Moreover, in addition to the one electron not included in the atomic


masses, a positron of mass equal to that of an electron is created. Consequently, one must add the rest energies of two electrons to the rest energy of the daughter atomic mass when calculating Q.

38 Why isn't there an element with Z = 130? An element with such a high Z value would either fission spontaneously or decay almost immediately by a emission (see Fig. 40 39 Why is a moderator needed in an ordinary nuclear fission reactor? The probability for neutron capture by the fissionable nucleus is large only for slow (thermal) neutrons. The neutrons emitted in the fission process are fast (high energy) neutrons and must be slowed to thermal neutrons before they are likely to be captured by another fissionable nucleus. 40 Explain why water is more effective than lead in slowing down fast neutrons. See text p. 235 and Figure 8-34. The process of slowing down involves the sharing of energy of a fast neutron and another nucleus in an elastic collision. The fast particle will lose maximum energy in such a collision if the target particle is of the same mass as the incident particle. Hence, neutron-proton collisions are most effective in slowing down neutrons. However, ordinary water cannot be used as a moderator because protons will capture the slow neutrons and form deuterons. 41* What happens to the neutrons produced in fission that do not produce another fission? Some of the neutrons are captured by other nuclei. Those not captured decay according to the reaction n p + e +

_ ne.

42 What is the advantage of a breeder reactor over an ordinary one? What are the disadvantages? Advantages: The reactor uses 238U which, by neutron capture and subsequent decays, produces 239Pu. This plutonium isotope fissions by fast neutron capture. Thus, the breeder reactor uses the plentiful uranium isotope and does not need a moderator to slow the neutrons needed for fission. Disadvantages: The fraction of delayed neutrons emitted in the fission of 239Pu is very small. Consequently, control of the fission reaction is very difficult, and the safety hazards are more severe than for the ordinary reactor that uses 235U as fuel. (See text, p. 1303.) 43 Assuming an average energy of 200 MeV per fission, calculate the number of fissions per second needed for a

500-MW reactor. 1. Convert 500 MW to MeV/s 2. N = P/(200 MeV)

5108 J/s = 3.1251021 MeV/s N = 1.561019 fissions/s

44 If the reproduction factor in a reactor is k = 1.1, find the number of generations needed for the power level to (a)

double, (b) increase by a factor of 10, and (c) increase by a factor of 100. Find the time needed in each case if (d) there are no delayed neutrons, so the time between generations is 1 ms, and (e) there are delayed neutrons that make the average time between generations 100 ms.

(d): (a), (b), (c) Proceed as in Example 40-7 (e): All times are increased by a factor of 100

(a) N = ln 2/ln 1.1 = 7.27; t = 7.27 ms; (b) t = 24.2 ms; (c) t = 48.3 ms (a) 0.727 s; (b) 2.42 s; (c) 4.83 s

45* Compute the temperature T for which kT = 10 keV, where k is Boltzmann's constant. k = 8.6210-5 eV/K = 8.6210-8 keV/K T = 10/8.6210-8 = 1.16108 K


46 In 1989, researchers claimed to have achieved fusion in an electrochemical cell at room temperature. They claimed a power output of 4 W from deuterium fusion reactions in the palladium electrode of their apparatus.

(a) If the two most likely reactions are 2H + 2H 3He + n + 3.27 MeV and 2H + 2H 3H + 1H + 4.03 MeV with 50% of the reactions going by each branch, how many neutrons per second would we expect to be emitted in the generation of 4 W of power? 1. Find the number of reactions per second 2. Number of neutrons emitted per second = N/2

N = [2(4/1.6 10-13)/7.30] s-1 = 6.841012 s-1. Nn = 3.421012 neutrons/s

47 A fusion reactor using only deuterium for fuel would have the two reactions in Problem 46 taking place in it. The

3H produced in the second reaction reacts immediately with another 2H to produce 3H + 2H 4He + n + 17.6 MeV The ratio of 2H to 1H atoms in naturally occurring hydrogen is 1.5 10 4. How much energy would be produced 1. Find the energy released using 5 2H nuclei 2. Find the number of H atoms in 4 L of H2O 3. Find the number of D atoms in 4 L of H2O 4. Find the energy produced

Q = (7.30 + 17.6) MeV = 24.9 MeV NH = 2(4/18)NA = 2.6761026 ND = NH1.510-4 = 4.011022 E = (ND/5)Q = 19.971022 MeV = 3.201010 J

48 The fusion reaction between 2H and 3H is 3H + 2H 4He + n + 17.6 MeV Using the given Q value, find the final energies of both the 4He nucleus and the neutron, assuming the initial kinetic energy of the system is 1.00 MeV and the initial momentum of the system is zero. 1. Write the condition for energy conservation 2. Write the condition for momentum conservation 3. Solve for KHe 4. Solve for Kn = 18.6 MeV - KHe

18.6 MeV = 1/2mHevHe2 + 1/2mnvn2 = KHe + Kn (1) vHemHe + vnmn = 0; vHe2 = vn2(mn/mHe)2 (2) KHe = (18.6 MeV)/(1 + mn/MHe) = 14.86 MeV Kn = 3.74 MeV

Note: In the first printing of the textbook the problem statement reads, The fusion reaction between 2H and 3H is 3H + 2H 4He + n + 17.6 MeV Using the conservation of momentum and the given Q value, find the final energies of both the 4He nucleus and the neutron, assuming that the initial momentum of the system is zero. This problem cannot be solved without additional information. For the 2H and 3H to interact, the nuclei must come within range of the nuclear forces. This requires a minimum of about 0.6 MeV in the center-of-mass reference frame. One must know the initial energy of the system to find KHe and Kn. 49* Energy is generated in the sun and other stars by fusion. One of the fusion cycles, the protonproton cycle, consists of the following reactions: 1H + 1H 2H + b + + ne 1H + 2H 3He + g


followed by 1H + 3He 4He + b + + ne (a) Show that the net effect of these reactions is 41H 4He + 2b + + 2ne + g (b) Show that rest energy of 24.7 MeV is released in this cycle (not counting the energy of 1.02 MeV released when each positron meets an electron and the two annihilate). (c) The sun radiates energy at the rate of about 41026 W. Assuming this is due to the conversion of four protons into helium plus g rays and neutrinos, which releases 26.7 MeV, what is the rate of proton consumption in the sun? How long will the sun last if it continues to radiate at its present level? (Assume that protons constitute about half of the total mass [2 1030 kg] of the sun.) (a) 1. Sum the three reactions 2. Cancel the common quantities (b) Dmc2 = (4mp - ma - 4me)c2 (see Problem 37) (c) 1. Find N, the number of protons in the sun 2. Find the energy released per proton in fusion 3. Find R, the rate of proton consumption 4. Find T, the time for consumption of all protons

4 1H + 2H + 3He 2H + 3He + 4He + 2b + + 2ne + g 4 1H 4He + 2b + + 2ne + g

Dmc2 = [(41.007825 - 4.002603)931.5 - 40.511] MeV = 25.69 MeV N = 1030/1.67310-27 = 5.981056 E = 26.7/4 MeV = 6.675 MeV = 1.0710-12 J R = P/E = 41026/1.0710-12 s-1 = 3.7451038 s-1 T = 5.981056/3.7451038 s = 1.601018 s = 5.051010 y

50 True or false: (a) The atomic nucleus contains protons, neutrons, and electrons. (b) The mass of 2H is less than the mass of a proton plus a neutron. (c) After two half-lives, all the radioactive nuclei in a given sample have decayed. (d) In a breeder reactor, fuel can be produced as fast as it is consumed. (a) False (b) True (c) False (d) True (given an unlimited supply of 238U) 51 Why do extreme changes in the temperature or pressure of a radioactive sample have little or no effect on the

radioactivity? Pressure and temperature changes have no effect on the internal structure of the nucleus. They do have an effect on the electronic configuration; consequently, they can influence K-capture processes. 52 The stable isotope of sodium is 23Na. What kind of radioactivity would you expect of (a) 22Na and (b) 24Na? (a) b + decay; (b) b - decay 53* Why does fusion occur spontaneously in the sun but not on earth? Fusion requires extremely high temperature and pressure. These conditions are met in the core of the sun but not on earth. 54 (a) Show that ke2 = 1.44 MeVfm, where k is the Coulomb constant and e is the electron charge. (b) Show that

hc = 1240 MeVfm.

(a) ke2 = [8.99109(1.610-19)2] J.m = [8.991091.610-25] MeV.m = 8.990.16 MeV.fm = 1.44 MeV.fm

(b) hc = 1240 eV.nm (see Equ. 17-5) = 1240 MeV.fm 55 The counting rate from a radioactive source is 6400 counts/s. The half-life of the source is 10 s. Make a plot of the

counting rate as a function of time for times up to 1 min. What is the decay constant for this source?


The decay constant is l = 0.693/t1/2 = 0.0693 s-1. The figure shows the function R = 6400e-0.0693t for 0 < t < 60 s.

56 Find the energy needed to remove a neutron from (a) 4He and (b) 7Li (a), (b) Use Equ. 40-19 and Table 40-1 (a) Dm = 0.022092 u; E = 20.98 MeV

(b) Dm = 0.007786 u; E = 7.253 MeV 57* The isotope 14C decays according to 14C 14N + e + n e . The atomic mass of 14N is 14.003074 u. Determine

the maximum kinetic energy of the electron. (Neglect recoil of the nitrogen atom.) Emax = Q = [m(14C) - m(14N)]c2 Emax = (0.003242 - 0.003074)931.5 MeV = 156.5 keV 58 A neutron star is an object of nuclear density. If our sun were to collapse to a neutron star, what would be the

radius of that object?

p?M

R = /

43

31

; r = 1.1741017 kg/m3 (see Prob. 7) km 15.9 = m1017414109913

17

30 31

..

R = /

p

59 Nucleus A has a half-life that is twice that of nucleus B. At t = 0 the number of B nuclei in a sample is twice that

of A nuclei. If the half-life of A is 1 h, will there ever be an instant when the number of A and B nuclei are equal? If so, when will this moment occur?

We are given that N0,B = 2N0,A and t1/2,A = 2t1/2,B. Therefore, lA = lB/2. At any time t, e N = t

A0,AAl-N and

eNeNN t2A0,tB0,B AB 2 = = ll -- . Thus, NB = NA when ee 22 tt AA = ll -- . Hence, t = (ln 2)/lA = t1/2,A = 1 h.

60 Calculate the nuclear radii of 19F, 145La, and 246Cm. (a), (b), (c) Use Equ. 40-1 (a) 4.00 fm; (b) 7.88 fm; (c) 9.40 fm 61* The relative abundance of 40K (molecular mass 40.0 g/mol) is 1.2 10-4. The isotope 40K is radioactive with a half-life of 1.3 109 y. Potassium is an essential element of every living cell. In the human body the mass of potassium


constitutes approximately 0.36% of the total mass. Determine the activity of this radioactive source in a student whose mass is 60 kg. 1. Find N, the number of K nuclei in the person 2. Find N40, the number of 40K nuclei 3. Find N40l, the activity of the 40K nuclei

N = 600.00366.021026/39.1 = 3.3261024 N40 = 1.210-4 3.3261024 = 3.991020

l = [0.693/(1.3 109 3.156107)] s-1 = 1.6910-17 s-1; activity = 6.74103 Bq

62 A 0.05394-kg sample of 144Nd (atomic mass 143.91 u) emits an average of 2.36 a particles each second. Find the

decay constant in s 1 and the half-life in years 1. Determine N0 2. l = R0/N0 (Equ. 40-8) 3. t1/2 = 0.693/l

N0 = (53.94/143.91)NA = 2.2561023

l = 2.36/2.2561023 = 1.04610-23 s-1 t1/2 = 6.6261022 s = 2.101015 y

63 The isotope 24Na is a b emitter with a half-life of 15 h. A saline solution containing this radioactive isotope with an

activity of 600 kBq is injected into the bloodstream of a patient. Ten hours later, the activity of 1 mL of blood from this individual yields a counting rate of 60 Bq. Determine the volume of blood in this patient.

1. Use Equ. 40-12 to find R0 of 1 mL of blood 2. Volume of blood = 6 105/95.24 mL

R0/mL = 6022/3 Bq = 95.24 Bq V = 6.30 L

64 Determine the closest distance of approach of an 8-MeV a particle in a head-on collision with a nucleus of 197Au

and a nucleus of 10B, (a) neglecting the recoil of the struck nuclei. (b) Repeat the calculation taking into account the recoil of the struck nuclei.

We shall first do this problem for the general case of an a particle in a head-on collision with a nucleus of atomic mass M u. In the CM frame, the kinetic energy is KCM = Klab/(1 + 4/M), where we have taken the atomic mass of the a

particle to be 4 u. At the point of closest approach, KCM = kq1q2/Rmin = (1.44 MeV.fm)(2Z)/Rmin. So we find that Rmin

= 2Z(1.44 MeV.fm)/KCM. Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab in the above expression for Rmin.

(a) 1. Find Rmin for 197Au 2. Find Rmin for 10B (b) 1. Find KCM for the case of 197Au 2. Find Rmin 3. Find KCM for the case of 10B; find Rmin

Rmin = (2791.44/8) fm = 28.44 fm Rmin = (251.44/8) fm = 1.8 fm KCM = (8/1.0203) MeV = 7.841 MeV Rmin = 29.02 fm (2% greater than without recoil) KCM = (8/1.40) MeV; Rmin = 2.52 fm (recoil is important)

65* Twelve nucleons are in a one-dimensional infinite square well of length L = 3 fm. (a) Using the approximation that the mass of a nucleon is 1 u, find the lowest energy of a nucleon in the well. Express your answer in MeV. What is the groundonly 2 in each state and (c) 6 of the nucleons are neutrons and 6 are protons so that there can be 4 nucleons in each state? (Neglect the energy of Coulomb repulsion of the protons.)


(a) Use Equ. 36-13 with n = 1

(b) Neutrons are fermions; only 2 per state;

En = n2E1

(c) Find E for 4 protons and 4 neutrons

MeV 22.96 = J109106618

1062663027

234

1 .).(

= E ---

0

E = 2(E1 + E2 + E3 + E4 + E5 + E6) = 29122.96 MeV = 4.178 GeV E = 4(E1 + E2 + E3) = 56E1 = 1.286 GeV

66 The helium nucleus or a particle is a very tightly bound system. Nuclei with N = Z = 2n, where n is an integer, such

as 12C, 16O, 20Ne, and 24Mg, may be thought of as agglomerates of a particles. (a) Use this model to estimate the binding energy of a pair of a particles from the atomic masses of 4He and 16O. Assume that the four a particles in 16O form a regular tetrahedron with one a particle at each vertex. (b) From the result obtained in part (a) determine, on the basis of this model, the binding energy of 12C and compare your result with that obtained from the atomic mass of 12C.

(a) 1. Find the binding energy for this model 2. For the tetrahedron there are 6 bonds (b) 1. 12C has 3 pairwise a particle bonds; find the total BE for 12C with this model 2. Use Table 40-1 to find BE( 12C)

BE = (44.002603 - 15.994915)uc2 = 0.015497 uc2 BE/bond = 0.002583 uc2 = 2.406 MeV BE( 12C) = 3BE( 4He) + 32.406 MeV; BE( 4He) = 28.3 MeV; BE( 12C) = 92.12 MeV BE( 12C) = (61.007825 + 6 1.008665 - 12.0000)uc2 = 92.16 MeV, in good agreement with the model

67 Radioactive nuclei with a decay constant of l are produced in an accelerator at a constant rate Rp. The number of

radioactive nuclei N then obeys the equation dN/dt = Rp lN. (a) If N is zero at t = 0, sketch N versus t for this situation. (b) The isotope 62Cu is produced at a rate of 100 per second by placing ordinary copper (63Cu) in a beam of high-energy photons. The reaction is

g + 63Cu 62Cu + n 62Cu decays by b decay with a half-life of 10 min. After a time long enough so that dN/dt 0, how many 62Cu nuclei are there?

(a) The solution to the differential equation dN/dt = Rp -

lN with the initial condition N(0) = 0 is N(t) = (Rp/l)(1 - e-lt ). A plot of this function is shown in the figure. Note that N(t) approaches Rp/l in the same manner that the charge on a capacitor approaches the value CV.

(b) 1. Find the decay constant l = (0.693/600) s-1 = 0.001155 s-1


2. Determine N N = 100/l = 8.658104 nuclei 68 The total energy consumed in the United States in 1 y is about 7.0 1019 J. How many kilograms of 235U would be

needed to provide this amount of energy if we assume that 200 MeV of energy is released by each fissioning uranium nucleus, that all of the uranium atoms undergo fission, and that all of the energy-conversion mechanisms used are 100% efficient?

1. Determine N, the number of fissions required 2. Find the mass of 235U required

N = 71019/(2001.610-13) = 2.191030 M235 = (2.191030/6.021026)235 kg = 8.54105 kg

69* (a) Find the wavelength of a particle in the ground state of a one-dimensional infinite square well of length L = 2 fm. (b) Find the momentum in units of MeV/c for a particle with this wavelength. (c) Show that the total energy of an

electron with this wavelength is approximately E pc. (d) What is the kinetic energy of an electron in the ground state f this well? This calculation shows that if an electron were confined in a region of space as small as a nucleus, it

would have a very large kinetic energy. (a) In ground state, l = 2L (see Equ. 17-17) (b) Use Equ. 17-7, p = h/l = hc/lc (c) E 2 = E02 + p2c2; E0 = 0.511 MeV


mass M that is at rest in the laboratory frame of reference. (a) Show that the speed of the center of mass in the lab frame is V = mvL/(m + M). (b) What is the speed of the nucleus in the center-of-mass frame before the collision? After the collision? (c) What is the speed of the nucleus in the lab frame after the collision? (d) Show that the energy of the nucleus after the collision in the lab frame is

mv)(m + M

mM = )V M( L22

2

214

221

(e) Show that the fraction of the energy lost by the neutron in this elastic collision is

) + m/M(

(m/M) =

)(m + MmM

= E

E22 1

44-

This Problem is essentially identical to Example 8-17, text p.235; replace m1 by m, m2 by M, v1i by vL, and v2f by vMf. We denote the CM velocities by capital letters. (Note: In the first printing, there are several misprints in Example 8-17.) (a) Use Equ. 8-7 (b) 1. In the CM frame, VMi = V 2. In the CM frame, Vf = -Vi (c) See Example 8-17 (d) KM = 1/2MvMf2 (e) DE = -KMf, E = 1/2mvL2

(m + M)V = mvL; V = mvL/(m + M) VMi = V VMf = -V vMf = 2mvL/(m + M) KMf = 2Mm2vL2/(m + M)2 -DE/E = 4mM/(m + M)2 = 4(m/M)/(1 + m/M)2

73 (a) Use the result of part (e) of Problem 72 (Equation 40-23) to show that after N head-on collisions of a neutron

with carbon nuclei at rest, the energy of the neutron is approximately (0.714)NE0, where E0 is its original energy. (b) How many head-on collisions are required to reduce the energy of the neutron from 2 MeV to 0.02 eV, assuming stationary carbon nuclei?

(a) 1. Determine f = Ef/E0 per collision; m = 1.008665u, M = 12.00000u 2. After N collisions, EfN = fNE0 (b) N = [ln (EfN/E0)]/ln(0.714)

f = (E0 - DE)/E0 = 1 - DE/E0 = 0.714 EfN = (0.714)NE0 N = ln(10-8 )/ln(0.714) = 54.7; 55 collisions

74 On the average, a neutron loses 63% of its energy in a collision with a hydrogen atom and 11% of its energy in a

collision with a carbon atom. Calculate the number of collisions needed to reduce the energy of a neutron from 2 MeV to 0.02 eV if the neutron collides with (a) hydrogen atoms and (b) carbon atoms. (See Problem 73.) Note the difference between the energy loss per collision specified here and that used in the preceding problem. In the preceding problem it was assumed that all collisions are head-on collisions.

(a) N = [ln (EfN/E0)]/ln(0.37); EfN/E0 = 10-8 (b) N = [ln (EfN/E0)]/ln(0.89); EfN/E0 = 10-8

N = 18.5; 19 collisions are required N = 158; 158 collisions are required


75 Frequently, the daughter of a radioactive parent is itself radioactive. Suppose the parent, designated by A has a

decay constant lA, while the daughter, designated B has a decay constant lB. The number of nuclei of B are then given by the solution to the differential equation

dNB/dt = lANA - lBNB (a) Justify this differential equation. (b) Show that the solution for this equation is

NB(t) = )e e(B tt

AB

AA BA ll

lll -- -

-0

where NA0 is the number of A nuclei present at t = 0 when there are no B nuclei. (c) Show that NB(t) > 0 whether lA > lB or lB > lA. (d) Make a plot of NA(t) and NB(t) as a function of time when tB = 3tA. (a) The rate of change of NB is the rate of generation of B nuclei minus the rate of decay of B nuclei. The generation rate is equal to the decay rate of A nuclei, which equals lANA. The decay rate of B nuclei is lBNB.

(b) From Equ. 40-6, e N = t0AA AlN . The differential equation for NB therefore is

N = BB0AAB A ll l --eN

dtdN t

. If we differentiate the expression for NB respect to t we obtain the

differential equation, which demonstrates that ( )

= )( BAAB

A0AB ee

NtN tt llll

l -- --

is the solution to the differential

equation for NB. (c) If lA > lB the denominator and the expression in the parentheses are both negative for t > 0. If lB > lA the

denominator and the expression in the parentheses are both positive for t > 0. (d) If tB = 3tA, lB = lA/3. The figure below shows NA(t) and NB(t).


76 Suppose isotope A decays to isotope B with a decay constant lA, and isotope B in turn decays with a decay

constant lB. Suppose a sample contains, at t = 0, only isotope A. Derive an expression for the time at which the number of isotope B nuclei will be a maximum. ( See Problem 75.)

Set dNB/dt = 0 in the expression given in Problem 40-75. Replace lANA by e N t0AA All - and NB by

( )

BA

AB

A0A eeN tt ll

lll -- -

-. Simplifying the equation one finds that the condition for dNB/dt = 0 is

e = e tAtB AB ll ll -- , or llll

AB

AB

)/ln

-(

= t .

77* An example of the situation discussed in Problem 75 is the radioactive isotope 229Th, an a emitter with a half-life of 7300 years. Its daughter, 225Ra, is a b emitter with a half-life of 14.8 d. In this, as in many instances, the half-life of the parent is much longer than that of the daughter. Using the expression given in Problem 75 (b), show that, starting with a sample of pure 229Th containing NA0 nuclei, the number, NB, of 225Ra nuclei will, after a several years, be a constant, given by NB = (lA/lB)NA The number of daughter nuclei are said to be in secular equilibrium.

Since tA >> tB, lA

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