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Ch40 ssm

M
Marta Díaz

The document provides information about calculating binding energies of nuclei from their masses. It gives the calculations for the binding energy and binding energy per nucleon for 12C, 56Fe, and 238U. The thermal energy of neutrons at 25°C is also calculated, as is the half-life of neutrons based on the reduction of intensity of a monoenergetic neutron beam over distance. Rutherford's hypothesis that electrons could exist within the nucleus is shown to be invalid due to the high kinetic energies electrons would have based on the Heisenberg uncertainty principle.

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Chapter 40 Nuclear Physics Properties of Nuclei 15 • Calculate the binding energy and the binding energy per nucleon from the masses given in Table 40-1 for (a) 12 C, (b) 56 Fe, and (c) 238 U. Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2 . To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 12 C, Z = 6 and N = 6. Add the mass of the neutrons to that of the protons: u940098.12u6651.0086u825007.1666 np =×+×=+ mm Subtract the mass of 12 C from this result: ( ) u9400.098u12u940098.1266 Cnp 12 =−=−+ mmm Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV2.92 u MeV/5.931 u940098.0 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV68.7 12 MeV2.92b == A E (b) For 56 Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the protons: u400463.56u6651.00803u825007.1263026 np =×+×=+ mm Subtract the mass of 56 Fe from this result: ( ) u4580.528u942934.55u400463.563026 Cnp 12 =−=−+ mmm 52
Nuclear Physics 53 Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV492 u MeV/5.931 u4580.528 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV79.8 56 MeV492b == A E (c) For 238 U, Z = 92 and N = 146. Add the mass of the neutrons to that of the protons: u990984.239u6651.008146u825007.19214692 np =×+×=+ mm Subtract the mass of 238 U from this result: ( ) u207934.1u783050.238u990984.23914692 Unp 238 =−=−+ mmm Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV1802 u MeV/5.931 u207934.1 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV57.7 238 MeV1802b == A E 19 •• The neutron, when isolated from an atomic nucleus, decays into a proton, an electron, and an antineutrino as follows: n → 1 H +e− +ν . The thermal energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In both joules and electron volts, calculate the energy of a thermal neutron at 25ºC. (b) What is the speed of this thermal neutron? (c) A beam of monoenergetic thermal neutrons is produced at 25ºC and has an intensity I. After traveling 1350 km, the beam has an intensity of 1 2 I. Using this information, estimate the half-life of the neutron. Express your answer in minutes. Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from I to I/2, is the half-life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed.
Chapter 4054 (a) The thermal energy of the neutron is: ( )( ) meV7.25 J101.602 eV1 J1011.4 J1011.4 K27325J/K10381.1 19 21 21 23 thermal = × ××= ×= +×= = − − − − kTE (b) Equate and the kinetic energy of the neutron to obtain: thermalE 2 n2 1 thermal vmE = ⇒ n thermal2 m E v = Substitute numerical values and evaluate v: ( ) km/s22.2 kg101.673 J104.112 27 21 = × × = − − v (c) Relate the half-life, ,21t to the speed of the neutrons in the beam: v x t =21 Substitute numerical values and evaluate :21t min1.10 s60 min1 s086 km/s2.22 km1350 1/2 = ×==t 21 •• In 1920, 12 years before the discovery of the neutron, Ernest Rutherford argued that proton–electron pairs might exist in the confines of the nucleus in order to explain the mass number, A, being greater than the nuclear charge, Z. He also used this argument to account for the source of beta particles in radioactive decay. Rutherford’s scattering experiments in 1910 showed that the nucleus had a diameter of approximately 10 fm. Using this nuclear diameter, the uncertainty principle, and that beta particles have an energy range of 0.02 MeV to 3.40 MeV, show why the hypothetical electrons cannot be confined to a region occupied by the nucleus. Picture the Problem The Heisenberg uncertainty principle relates the uncertainty in position, Δx, to the uncertainty in momentum, Δp, by ΔxΔp ≥ 1 2 h. Solve the Heisenberg equation for Δp: x p Δ ≈Δ 2 h
Nuclear Physics 55 Assuming that electrons are contained within the nucleus, the uncertainty in their momenta must be: ( ) m/skg1025.5 m10102 sJ1005.1 21 15 34 ⋅×= × ⋅× ≈Δ − − − p The kinetic energy of the electron is given by: pcK = Substitute numerical values and evaluate K for a nuclear electron: ( )( ) MeV88.9 J101.602 eV1 J1058.1m/s10998.2m/skg1025.5 19 12821 = × ××=×⋅×= − −− K This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of 1 to 1000 eV. Hence the premise on which it is based (that electrons are contained within the nucleus) must be false. Radioactivity 29 •• Plutonium is very toxic to the human body. Once it enters the body it collects primarily in the bones, although it also can be found in other organs. Red blood cells are synthesized within the marrow of the bones. The isotope 239 Pu is an alpha emitter that has a half-life of 24 360 years. Because alpha particles are an ionizing radiation, the blood-making ability of the marrow is, in time, destroyed by the presence of 239 Pu. In addition, many kinds of cancers will also develop in the surrounding tissues because of the ionizing effects of the alpha particles. (a) If a person accidentally ingested 2.0 μg of 239 Pu and all of it is absorbed by the bones of the person, how many alpha particles are produced per second within the body of the person? (b) When, in years, will the activity be 1000 alpha particles per second? Picture the Problem Each 239 Pu nucleus emits an alpha particle whose activity, A, depends on the decay constant of 239 Pu and on the number N of nuclei present in the ingested 239 Pu. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of 239 Pu. Finally, we can use the dependence of the activity on time to find the time at which the activity be 1000 alpha particles per second. (a) The activity of the nuclei present in the ingested 239 Pu is given by: NA λ= (1)
Chapter 4056 Find the constant for the decay of 239 Pu: ( ) ( )( ) 113 2/1 s1002.9 Ms/y56.31y24360 693.02ln −− ×= == t λ Express the number of nuclei N present in the mass mPu of 239 Pu ingested: Pu Pu A M m N N = ⇒ Pu A Pu M N mN = where MPu is the atomic mass of 239 Pu. Substitute numerical values and evaluate N: ( ) nuclei1004.5 g/mol239 nuclei/mol1002.6 g0.2 15 23 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = μN Substitute numerical values in equation (1) and evaluate A: ( )( ) /s1055.4 1004.5s1002.9 3 15113 α α ×= ××= −− A (b) The activity varies with time according to: t eAA λ− = 0 ⇒ λ− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 0 ln A A t Substitute numerical values and evaluate t: ( ) y1032.5 y Ms56.31 s1002.9 s/1055.4 s/1000 ln 4 113 3 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×− ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = −− α α t 31 • The fissile material 239 Pu is an alpha emitter. Write the reaction that describes 239 Pu undergoing alpha decay. Given that 239 Pu, 235 U, and an alpha particle have respective masses of 239.052 156 u, 235.043 923 u, and 4.002 603 u, use the relations appearing in Problem 32 to calculate the kinetic energies of the alpha particle and the recoiling daughter nucleus. Picture the Problem We can write the equation of the decay process by using the fact that the post-decay sum of the Z and A numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations from Problem 32 is given by Q = −(Δm)c2 . 239 Pu undergoes alpha decay according to: Q++→ α4 2 235 92 239 94 UPu
Nuclear Physics 57 The Q value for the decay is given by: ( ) ( )[ ] ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= u MeV/931.5 2 2 UPu c cmmmQ α Substitute numerical values and evaluate Q: ( ) ( )[ ] MeV24.5 u 931.5MeV/ u4.002603u235.043923u239.052156 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= c cQ From Problem 32, the kinetic energy of the alpha particle is given by: Q A A K ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 4 α Substitute numerical values and evaluate Kα: ( ) MeV5.15 MeV5.24 239 4239 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =αK From Problem 32, the kinetic energy of the 235 U is given by: A Q K 4 U = Substitute numerical values and evaluate :U235K ( ) keV2.89 235 MeV24.54 U235 ==K 35 •• Radiation has been used for a long time in medical therapy to control the development and growth of cancer cells. Cobalt-60, a gamma emitter that emits photons that have energies of 1.17 MeV and 1.33 MeV, is used to irradiate and destroy deep-rooted cancers. Small needles made of 60 Co of a specified activity are encased in gold and used as body implants in tumors for time periods that are related to tumor size, tumor cell reproductive rate, and the activity of the needle. (a) A 1.00 μg sample of 60 Co, that has a half-life of 5.27 y, that is used to irradiate a small internal tumor with gamma rays, is prepared in the cyclotron of a medical center. Determine the activity of the sample in curies. (b) What will the activity of the sample be 1.75 y from now? Picture the Problem We can use NR λ=0 to find the initial activity of the sample and to find the activity of the sample after 1.75 y.t eRR λ− = o
Chapter 4058 (a) The initial activity of the sample is the product of the decay constant λ for 60 Co and the number of atoms N of 60 Co initially present in the sample: NR λ=0 Express N in terms of the mass m of the sample, the molar mass M of 60 Co, and Avogadro’s number NA: AN M m N = Substituting for N yields: M mN R A 0 λ = The decay constant is given by: ( ) 1/2 2ln t =λ Substitute for λ to obtain: ( ) Mt mN R 1/2 A 0 2ln = Substitute numerical values and evaluate R0: ( )( ) mCi13.1 s103.7 Ci1 s10183.4 mol g 60 y Ms31.56 y27.5 mol nuclei 106.022g101.002ln 110 17 236 0 = × ××= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×× = − − − R (b) The activity varies with time according to: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − == 5.27y 0.693 oo t t eReRR λ Evaluate R at t = 1.75 y: ( ) mCi898.0 mCi1.13 5.27y 1.75y0.693 = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × − eR Nuclear Reactions 43 •• (a) Use the values 14.003 242 u and 14.003 074 u for the atomic masses of and 14 , respectively, to calculate the Q value (in MeV) for the β- decay reaction 14 C N 6 14 C → 7 14 N + e− + ve . (b) Explain why you should not add the mass of the electron to that of atomic 14 for the calculation in (a).N
Nuclear Physics 59 Picture the Problem We can use ( ) 2 cmQ Δ−= to find the Q values for this reaction. (a) The masses of the atoms are: u242003.14C14 =m u074003.14N14 =m Calculate the increase in mass: u168000.0 u242003.14u074003.14 if −= −= −=Δ mmm Calculate the Q value: ( ) ( ) MeV156.0 u MeV/ 5.931u168000.0 2 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= Δ−= c c cmQ (b) The masses given are for atoms, not nuclei, so for nuclear masses the masses are too large by the atomic number multiplied by the mass of an electron. For the given nuclear reaction, the mass of the carbon atom is too large by and the mass of the nitrogen atom is too large by . Subtracting from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron) is mathematically equivalent to explicitly subtracting from the right side of the equation. e6m e7m e6m e1m Fission and Fusion 45 • Assuming an average energy of 200 MeV per fission, calculate the number of fissions per second needed for a 500-MW reactor. Picture the Problem The power output of the reactor is the product of the number of fissions per second and energy liberated per fission. Express the required number N of fissions per second in terms of the power output P and the energy released per fission Eper fission: fissionperE P N =
Chapter 4060 Substitute numerical values and evaluate N: 119 19 8 s1056.1 MeV200 J101.602 eV1 s J 1000.5 MeV200 MW500 − − ×= × ×× = =N 47 •• Consider the following fission reaction: 92 . The masses of the neutron,235 U +n → 42 95 Mo+ 57 139 La +2n + Q 235 U, 95 Mo, and 139 La are 1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u, respectively. Calculate the Q value, in MeV, for this fission reaction. Picture the Problem We can use Q = −(Δm)c2 , where Δm = mf − mi, to calculate the Q value. The Q value, in MeV, is given by: ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Δ−= u MeV/5.931 2 2 c cmQ Calculate the change in mass Δm: ( ) ( ) u068223.0 u6651.008u923235.043 u6651.0082u348138.906u842905.94 if −= +− ++= −=Δ mmm Substitute for Δm and evaluate Q: ( ) MeV208 u MeV/931.5 u068223.0 2 2 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= c cQ General Problems 53 • 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? Picture the Problem We can use the given information regarding the half-life of the source to find its decay constant. We can then plot a graph of the counting rate as a function of time.
Nuclear Physics 61 The decay constant is related to the half-life of the source: ( ) ( ) 1 1 21 s069.0 s0693.0 s10 2ln2ln − − = === t λ The activity of the source is given by: ( ) ( )tt eeRR 1 s0693.0 0 Bq6400 − −− == λ The following graph of ( ) ( )t eR 1 s0693.0 Bq6400 − − = was plotted using a spreadsheet program. 0 1000 2000 3000 4000 5000 6000 7000 0 10 20 30 40 50 60 t (s) R(Bq) 57 •• Show that the 109 Ag nucleus is stable and does not undergo alpha decay, . The mass of the47 109 Ag → 2 4 He + 45 105 Rh + Q 109 Ag nucleus is 108.904 756 u, and the products of the decay are 4.002 603 u and 104.905 250 u, respectively. Picture the Problem We can show that 109 Ag is stable against alpha decay by demonstrating that its Q value is negative. The Q value, in MeV, for this reaction is: ( )[ ] ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+−= u MeV/ 5.931 2 2 AgRh c cmmmQ α Substitute numerical values and evaluate Q: ( )[ ] MeV88.2 u MeV/ 5.931u756904.108u250905.104u603002.4 2 2 −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+−= c cQ
Chapter 4062 Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, . Kinetic energy cannot be negative; hence, alpha decay cannot occur unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, . Alpha decay cannot take place unless the total rest mass decreases. DKKQ α += DP mmm α +> 69 ••• Assume that a neutron decays into a proton and an electron without the emission of a neutrino. The energy shared by the proton and the electron is then 0.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the proton must be equal and opposite the momentum of the electron. This determines the ratio of the energies of the two particles, but because the electron is relativistic, the exact calculation of these relative energies is somewhat challenging. (a) Assume that the kinetic energy of the electron is 0.782 MeV and calculate the momentum p of the electron in units of MeV/c. Hint: Use (Equation 39-27). (b) Using your result from Part (a), calculate the kinetic energy p ( )22222 mccpE += 2 /2mp of the proton. (c) Because the total energy of the electron and the proton is 0.782 MeV, the calculation in Part (b) gives a correction to the assumption that the energy of the electron is 0.782 MeV. What percentage of 0.782 MeV is this correction? Picture the Problem The momentum of the electron is related to its total energy through and its total relativistic energy E is the sum of its kinetic and rest energies. 2 0 222 EcpE += (a) Relate the total energy of the electron to its momentum and rest energy: 2 0 222 EcpE += (1) The total relativistic energy E of the electron is the sum of its kinetic energy and its rest energy: 0EKE += Substitute for E in equation (1) to obtain: ( ) 2 0 222 0 EcpEK +=+ Solving for p gives: ( ) c EKK p 02+ =
Nuclear Physics 63 Substitute numerical values and evaluate p: ( )( ) c c p MeV/19.1 MeV511.02MeV782.0MeV782.0 = ×+ = (b) Because pp = −pe: p 2 p p 2m p K = Substitute numerical values (see Table 7-1 for the rest energy of a proton) and evaluate Kp: ( ) ( ) eV752 MeV/28.9382 MeV/188.1 2 2 p == c c K (c) The percent correction is: %0962.0 MeV0.782 eV752p == K K 73 ••• Frequently, the daughter nucleus of a radioactive parent nucleus is itself radioactive. Suppose the parent nucleus, designated by P, has a decay constant λP; while the daughter nucleus, designated by D, has a decay constant λD. The number of daughter nuclei ND are then given by the solution to the differential equation dND/dt = λPNP – λDND where NP is the number of parent nuclei. (a) Justify this differential equation. (b) Show that the solution for this equation is N D t( )= N P0λP λD − λP e−λP t − e−λDt ( ) where NP0 is the number of parent nuclei present at t = 0 when there are no daughter nuclei. (c) Show that the expression for ND in Part (b) gives ND (t) > 0 whether λP > λD or λD > λP. (d) Make a plot of NP (t) and ND (t) as a function of time when τD = 3τP, where τD and τP are the mean lifetimes of the daughter and parent nuclei, respectively. Picture the Problem We can differentiate ( ) ( )tt ee N tN DP PD PP0 D λλ λλ λ −− − − = with respect to t to show that it is the solution to the differential equation dND/dt = λPNP − λDND.
Chapter 4064 (a) The rate of change of ND is the rate of generation of D nuclei minus the rate of decay of D nuclei. The generation rate is equal to the decay rate of P nuclei, which equals λPNP. The decay rate of D nuclei is λDND. (b) We’re given that: DDPP D NN dt dN λλ −= (1) ( ) ( )tt ee N tN DP PD PP0 D λλ λλ λ −− − − = (2) t eNN P P0P λ− = (3) Differentiate equation (2) with respect to t to obtain: ( )[ ] ( )[ ] [ ]tttt ee N ee dt dN tN dt d DPDP DP PD PP0 PD PP0 D λλλλ λλ λλ λ λλ λ −−−− +− − =− − = Substitute this derivative in equation (1) to obtain: [ ] ( )⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − −=+− − −−−−− ttttt ee N eNee N DPPDP PD PP0 DP0PDP PD PP0 λλλλλ λλ λ λλλλ λλ λ Multiply both sides by PD PD λλ λλ − and simplify to obtain: [ ] ( ) [ ]tt tt tttt ttttt ee N eNe N eNeNe N eN eeNeNee N DP DP DPPP DPPDP DP D P0 P0 D PP0 P0P0 D PP0 P0 P0P0 D PD DP D P0 λλ λλ λλλλ λλλλλ λλ λ λ λ λ λ λ λλ λλ λ −− −− −−−− −−−−− +−= +−= +−−= −− − =+− which is an identity and confirms that equation (2) is the solution to equation (1). (c) If λP > λD the denominator and expression in parentheses are both negative for t > 0. If λP < λD the denominator and expression in parentheses are both positive for t > 0.
Nuclear Physics 65 (d) The following graph was plotted using a spreadsheet program. 0.00 0 5 10 15 20 Number of parent (P) nuclei Number of daughter ( D) nuclei τ τ τ τp p p p NPO N N N NPO PO PO PO 0.20 0.40 0.60 0.80 1.00 Time

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Ch40 ssm

  • 1. Chapter 40 Nuclear Physics Properties of Nuclei 15 • Calculate the binding energy and the binding energy per nucleon from the masses given in Table 40-1 for (a) 12 C, (b) 56 Fe, and (c) 238 U. Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2 . To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 12 C, Z = 6 and N = 6. Add the mass of the neutrons to that of the protons: u940098.12u6651.0086u825007.1666 np =×+×=+ mm Subtract the mass of 12 C from this result: ( ) u9400.098u12u940098.1266 Cnp 12 =−=−+ mmm Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV2.92 u MeV/5.931 u940098.0 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV68.7 12 MeV2.92b == A E (b) For 56 Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the protons: u400463.56u6651.00803u825007.1263026 np =×+×=+ mm Subtract the mass of 56 Fe from this result: ( ) u4580.528u942934.55u400463.563026 Cnp 12 =−=−+ mmm 52
  • 2. Nuclear Physics 53 Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV492 u MeV/5.931 u4580.528 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV79.8 56 MeV492b == A E (c) For 238 U, Z = 92 and N = 146. Add the mass of the neutrons to that of the protons: u990984.239u6651.008146u825007.19214692 np =×+×=+ mm Subtract the mass of 238 U from this result: ( ) u207934.1u783050.238u990984.23914692 Unp 238 =−=−+ mmm Multiply the mass difference by c2 and convert to MeV: ( ) ( ) MeV1802 u MeV/5.931 u207934.1 2 22 b =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ= c ccmE and the binding energy per nucleon is MeV57.7 238 MeV1802b == A E 19 •• The neutron, when isolated from an atomic nucleus, decays into a proton, an electron, and an antineutrino as follows: n → 1 H +e− +ν . The thermal energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In both joules and electron volts, calculate the energy of a thermal neutron at 25ºC. (b) What is the speed of this thermal neutron? (c) A beam of monoenergetic thermal neutrons is produced at 25ºC and has an intensity I. After traveling 1350 km, the beam has an intensity of 1 2 I. Using this information, estimate the half-life of the neutron. Express your answer in minutes. Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from I to I/2, is the half-life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed.
  • 3. Chapter 4054 (a) The thermal energy of the neutron is: ( )( ) meV7.25 J101.602 eV1 J1011.4 J1011.4 K27325J/K10381.1 19 21 21 23 thermal = × ××= ×= +×= = − − − − kTE (b) Equate and the kinetic energy of the neutron to obtain: thermalE 2 n2 1 thermal vmE = ⇒ n thermal2 m E v = Substitute numerical values and evaluate v: ( ) km/s22.2 kg101.673 J104.112 27 21 = × × = − − v (c) Relate the half-life, ,21t to the speed of the neutrons in the beam: v x t =21 Substitute numerical values and evaluate :21t min1.10 s60 min1 s086 km/s2.22 km1350 1/2 = ×==t 21 •• In 1920, 12 years before the discovery of the neutron, Ernest Rutherford argued that proton–electron pairs might exist in the confines of the nucleus in order to explain the mass number, A, being greater than the nuclear charge, Z. He also used this argument to account for the source of beta particles in radioactive decay. Rutherford’s scattering experiments in 1910 showed that the nucleus had a diameter of approximately 10 fm. Using this nuclear diameter, the uncertainty principle, and that beta particles have an energy range of 0.02 MeV to 3.40 MeV, show why the hypothetical electrons cannot be confined to a region occupied by the nucleus. Picture the Problem The Heisenberg uncertainty principle relates the uncertainty in position, Δx, to the uncertainty in momentum, Δp, by ΔxΔp ≥ 1 2 h. Solve the Heisenberg equation for Δp: x p Δ ≈Δ 2 h
  • 4. Nuclear Physics 55 Assuming that electrons are contained within the nucleus, the uncertainty in their momenta must be: ( ) m/skg1025.5 m10102 sJ1005.1 21 15 34 ⋅×= × ⋅× ≈Δ − − − p The kinetic energy of the electron is given by: pcK = Substitute numerical values and evaluate K for a nuclear electron: ( )( ) MeV88.9 J101.602 eV1 J1058.1m/s10998.2m/skg1025.5 19 12821 = × ××=×⋅×= − −− K This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of 1 to 1000 eV. Hence the premise on which it is based (that electrons are contained within the nucleus) must be false. Radioactivity 29 •• Plutonium is very toxic to the human body. Once it enters the body it collects primarily in the bones, although it also can be found in other organs. Red blood cells are synthesized within the marrow of the bones. The isotope 239 Pu is an alpha emitter that has a half-life of 24 360 years. Because alpha particles are an ionizing radiation, the blood-making ability of the marrow is, in time, destroyed by the presence of 239 Pu. In addition, many kinds of cancers will also develop in the surrounding tissues because of the ionizing effects of the alpha particles. (a) If a person accidentally ingested 2.0 μg of 239 Pu and all of it is absorbed by the bones of the person, how many alpha particles are produced per second within the body of the person? (b) When, in years, will the activity be 1000 alpha particles per second? Picture the Problem Each 239 Pu nucleus emits an alpha particle whose activity, A, depends on the decay constant of 239 Pu and on the number N of nuclei present in the ingested 239 Pu. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of 239 Pu. Finally, we can use the dependence of the activity on time to find the time at which the activity be 1000 alpha particles per second. (a) The activity of the nuclei present in the ingested 239 Pu is given by: NA λ= (1)
  • 5. Chapter 4056 Find the constant for the decay of 239 Pu: ( ) ( )( ) 113 2/1 s1002.9 Ms/y56.31y24360 693.02ln −− ×= == t λ Express the number of nuclei N present in the mass mPu of 239 Pu ingested: Pu Pu A M m N N = ⇒ Pu A Pu M N mN = where MPu is the atomic mass of 239 Pu. Substitute numerical values and evaluate N: ( ) nuclei1004.5 g/mol239 nuclei/mol1002.6 g0.2 15 23 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = μN Substitute numerical values in equation (1) and evaluate A: ( )( ) /s1055.4 1004.5s1002.9 3 15113 α α ×= ××= −− A (b) The activity varies with time according to: t eAA λ− = 0 ⇒ λ− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 0 ln A A t Substitute numerical values and evaluate t: ( ) y1032.5 y Ms56.31 s1002.9 s/1055.4 s/1000 ln 4 113 3 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×− ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = −− α α t 31 • The fissile material 239 Pu is an alpha emitter. Write the reaction that describes 239 Pu undergoing alpha decay. Given that 239 Pu, 235 U, and an alpha particle have respective masses of 239.052 156 u, 235.043 923 u, and 4.002 603 u, use the relations appearing in Problem 32 to calculate the kinetic energies of the alpha particle and the recoiling daughter nucleus. Picture the Problem We can write the equation of the decay process by using the fact that the post-decay sum of the Z and A numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations from Problem 32 is given by Q = −(Δm)c2 . 239 Pu undergoes alpha decay according to: Q++→ α4 2 235 92 239 94 UPu
  • 6. Nuclear Physics 57 The Q value for the decay is given by: ( ) ( )[ ] ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= u MeV/931.5 2 2 UPu c cmmmQ α Substitute numerical values and evaluate Q: ( ) ( )[ ] MeV24.5 u 931.5MeV/ u4.002603u235.043923u239.052156 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= c cQ From Problem 32, the kinetic energy of the alpha particle is given by: Q A A K ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 4 α Substitute numerical values and evaluate Kα: ( ) MeV5.15 MeV5.24 239 4239 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =αK From Problem 32, the kinetic energy of the 235 U is given by: A Q K 4 U = Substitute numerical values and evaluate :U235K ( ) keV2.89 235 MeV24.54 U235 ==K 35 •• Radiation has been used for a long time in medical therapy to control the development and growth of cancer cells. Cobalt-60, a gamma emitter that emits photons that have energies of 1.17 MeV and 1.33 MeV, is used to irradiate and destroy deep-rooted cancers. Small needles made of 60 Co of a specified activity are encased in gold and used as body implants in tumors for time periods that are related to tumor size, tumor cell reproductive rate, and the activity of the needle. (a) A 1.00 μg sample of 60 Co, that has a half-life of 5.27 y, that is used to irradiate a small internal tumor with gamma rays, is prepared in the cyclotron of a medical center. Determine the activity of the sample in curies. (b) What will the activity of the sample be 1.75 y from now? Picture the Problem We can use NR λ=0 to find the initial activity of the sample and to find the activity of the sample after 1.75 y.t eRR λ− = o
  • 7. Chapter 4058 (a) The initial activity of the sample is the product of the decay constant λ for 60 Co and the number of atoms N of 60 Co initially present in the sample: NR λ=0 Express N in terms of the mass m of the sample, the molar mass M of 60 Co, and Avogadro’s number NA: AN M m N = Substituting for N yields: M mN R A 0 λ = The decay constant is given by: ( ) 1/2 2ln t =λ Substitute for λ to obtain: ( ) Mt mN R 1/2 A 0 2ln = Substitute numerical values and evaluate R0: ( )( ) mCi13.1 s103.7 Ci1 s10183.4 mol g 60 y Ms31.56 y27.5 mol nuclei 106.022g101.002ln 110 17 236 0 = × ××= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×× = − − − R (b) The activity varies with time according to: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − == 5.27y 0.693 oo t t eReRR λ Evaluate R at t = 1.75 y: ( ) mCi898.0 mCi1.13 5.27y 1.75y0.693 = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × − eR Nuclear Reactions 43 •• (a) Use the values 14.003 242 u and 14.003 074 u for the atomic masses of and 14 , respectively, to calculate the Q value (in MeV) for the β- decay reaction 14 C N 6 14 C → 7 14 N + e− + ve . (b) Explain why you should not add the mass of the electron to that of atomic 14 for the calculation in (a).N
  • 8. Nuclear Physics 59 Picture the Problem We can use ( ) 2 cmQ Δ−= to find the Q values for this reaction. (a) The masses of the atoms are: u242003.14C14 =m u074003.14N14 =m Calculate the increase in mass: u168000.0 u242003.14u074003.14 if −= −= −=Δ mmm Calculate the Q value: ( ) ( ) MeV156.0 u MeV/ 5.931u168000.0 2 2 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= Δ−= c c cmQ (b) The masses given are for atoms, not nuclei, so for nuclear masses the masses are too large by the atomic number multiplied by the mass of an electron. For the given nuclear reaction, the mass of the carbon atom is too large by and the mass of the nitrogen atom is too large by . Subtracting from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron) is mathematically equivalent to explicitly subtracting from the right side of the equation. e6m e7m e6m e1m Fission and Fusion 45 • Assuming an average energy of 200 MeV per fission, calculate the number of fissions per second needed for a 500-MW reactor. Picture the Problem The power output of the reactor is the product of the number of fissions per second and energy liberated per fission. Express the required number N of fissions per second in terms of the power output P and the energy released per fission Eper fission: fissionperE P N =
  • 9. Chapter 4060 Substitute numerical values and evaluate N: 119 19 8 s1056.1 MeV200 J101.602 eV1 s J 1000.5 MeV200 MW500 − − ×= × ×× = =N 47 •• Consider the following fission reaction: 92 . The masses of the neutron,235 U +n → 42 95 Mo+ 57 139 La +2n + Q 235 U, 95 Mo, and 139 La are 1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u, respectively. Calculate the Q value, in MeV, for this fission reaction. Picture the Problem We can use Q = −(Δm)c2 , where Δm = mf − mi, to calculate the Q value. The Q value, in MeV, is given by: ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Δ−= u MeV/5.931 2 2 c cmQ Calculate the change in mass Δm: ( ) ( ) u068223.0 u6651.008u923235.043 u6651.0082u348138.906u842905.94 if −= +− ++= −=Δ mmm Substitute for Δm and evaluate Q: ( ) MeV208 u MeV/931.5 u068223.0 2 2 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= c cQ General Problems 53 • 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? Picture the Problem We can use the given information regarding the half-life of the source to find its decay constant. We can then plot a graph of the counting rate as a function of time.
  • 10. Nuclear Physics 61 The decay constant is related to the half-life of the source: ( ) ( ) 1 1 21 s069.0 s0693.0 s10 2ln2ln − − = === t λ The activity of the source is given by: ( ) ( )tt eeRR 1 s0693.0 0 Bq6400 − −− == λ The following graph of ( ) ( )t eR 1 s0693.0 Bq6400 − − = was plotted using a spreadsheet program. 0 1000 2000 3000 4000 5000 6000 7000 0 10 20 30 40 50 60 t (s) R(Bq) 57 •• Show that the 109 Ag nucleus is stable and does not undergo alpha decay, . The mass of the47 109 Ag → 2 4 He + 45 105 Rh + Q 109 Ag nucleus is 108.904 756 u, and the products of the decay are 4.002 603 u and 104.905 250 u, respectively. Picture the Problem We can show that 109 Ag is stable against alpha decay by demonstrating that its Q value is negative. The Q value, in MeV, for this reaction is: ( )[ ] ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+−= u MeV/ 5.931 2 2 AgRh c cmmmQ α Substitute numerical values and evaluate Q: ( )[ ] MeV88.2 u MeV/ 5.931u756904.108u250905.104u603002.4 2 2 −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+−= c cQ
  • 11. Chapter 4062 Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, . Kinetic energy cannot be negative; hence, alpha decay cannot occur unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, . Alpha decay cannot take place unless the total rest mass decreases. DKKQ α += DP mmm α +> 69 ••• Assume that a neutron decays into a proton and an electron without the emission of a neutrino. The energy shared by the proton and the electron is then 0.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the proton must be equal and opposite the momentum of the electron. This determines the ratio of the energies of the two particles, but because the electron is relativistic, the exact calculation of these relative energies is somewhat challenging. (a) Assume that the kinetic energy of the electron is 0.782 MeV and calculate the momentum p of the electron in units of MeV/c. Hint: Use (Equation 39-27). (b) Using your result from Part (a), calculate the kinetic energy p ( )22222 mccpE += 2 /2mp of the proton. (c) Because the total energy of the electron and the proton is 0.782 MeV, the calculation in Part (b) gives a correction to the assumption that the energy of the electron is 0.782 MeV. What percentage of 0.782 MeV is this correction? Picture the Problem The momentum of the electron is related to its total energy through and its total relativistic energy E is the sum of its kinetic and rest energies. 2 0 222 EcpE += (a) Relate the total energy of the electron to its momentum and rest energy: 2 0 222 EcpE += (1) The total relativistic energy E of the electron is the sum of its kinetic energy and its rest energy: 0EKE += Substitute for E in equation (1) to obtain: ( ) 2 0 222 0 EcpEK +=+ Solving for p gives: ( ) c EKK p 02+ =
  • 12. Nuclear Physics 63 Substitute numerical values and evaluate p: ( )( ) c c p MeV/19.1 MeV511.02MeV782.0MeV782.0 = ×+ = (b) Because pp = −pe: p 2 p p 2m p K = Substitute numerical values (see Table 7-1 for the rest energy of a proton) and evaluate Kp: ( ) ( ) eV752 MeV/28.9382 MeV/188.1 2 2 p == c c K (c) The percent correction is: %0962.0 MeV0.782 eV752p == K K 73 ••• Frequently, the daughter nucleus of a radioactive parent nucleus is itself radioactive. Suppose the parent nucleus, designated by P, has a decay constant λP; while the daughter nucleus, designated by D, has a decay constant λD. The number of daughter nuclei ND are then given by the solution to the differential equation dND/dt = λPNP – λDND where NP is the number of parent nuclei. (a) Justify this differential equation. (b) Show that the solution for this equation is N D t( )= N P0λP λD − λP e−λP t − e−λDt ( ) where NP0 is the number of parent nuclei present at t = 0 when there are no daughter nuclei. (c) Show that the expression for ND in Part (b) gives ND (t) > 0 whether λP > λD or λD > λP. (d) Make a plot of NP (t) and ND (t) as a function of time when τD = 3τP, where τD and τP are the mean lifetimes of the daughter and parent nuclei, respectively. Picture the Problem We can differentiate ( ) ( )tt ee N tN DP PD PP0 D λλ λλ λ −− − − = with respect to t to show that it is the solution to the differential equation dND/dt = λPNP − λDND.
  • 13. Chapter 4064 (a) The rate of change of ND is the rate of generation of D nuclei minus the rate of decay of D nuclei. The generation rate is equal to the decay rate of P nuclei, which equals λPNP. The decay rate of D nuclei is λDND. (b) We’re given that: DDPP D NN dt dN λλ −= (1) ( ) ( )tt ee N tN DP PD PP0 D λλ λλ λ −− − − = (2) t eNN P P0P λ− = (3) Differentiate equation (2) with respect to t to obtain: ( )[ ] ( )[ ] [ ]tttt ee N ee dt dN tN dt d DPDP DP PD PP0 PD PP0 D λλλλ λλ λλ λ λλ λ −−−− +− − =− − = Substitute this derivative in equation (1) to obtain: [ ] ( )⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − −=+− − −−−−− ttttt ee N eNee N DPPDP PD PP0 DP0PDP PD PP0 λλλλλ λλ λ λλλλ λλ λ Multiply both sides by PD PD λλ λλ − and simplify to obtain: [ ] ( ) [ ]tt tt tttt ttttt ee N eNe N eNeNe N eN eeNeNee N DP DP DPPP DPPDP DP D P0 P0 D PP0 P0P0 D PP0 P0 P0P0 D PD DP D P0 λλ λλ λλλλ λλλλλ λλ λ λ λ λ λ λ λλ λλ λ −− −− −−−− −−−−− +−= +−= +−−= −− − =+− which is an identity and confirms that equation (2) is the solution to equation (1). (c) If λP > λD the denominator and expression in parentheses are both negative for t > 0. If λP < λD the denominator and expression in parentheses are both positive for t > 0.
  • 14. Nuclear Physics 65 (d) The following graph was plotted using a spreadsheet program. 0.00 0 5 10 15 20 Number of parent (P) nuclei Number of daughter ( D) nuclei τ τ τ τp p p p NPO N N N NPO PO PO PO 0.20 0.40 0.60 0.80 1.00 Time