When the spin component is antiparallel to the field the interaction energy is U =+ μzb.

Transcription

1 NUCLEAR PHYSICS 4 4 (a) 8 4 Si has 4 protons and 4 neutrons (b) 85 Rb has protons and 48 neutrons (c) 5 8 Tl has 8 protons and 4 neutrons 4 (a) Using R = ( fm) A, the radii are roughly fm, 5 fm, and fm (b) Using 4π R for each of the radii in part (a), the areas are fm, 5 fm and fm 4 (c) R fm, 4 fm and 499 fm (d) The density is the same, since the volume and the mass are both proportional to A: kg m (see Example 4) 44 (e) Dividing the result of part (d) by the mass of a nucleon, the number density is 4 fm = 4 m 4 IDENTIFY: Calculate the spin magnetic energy shift for each spin state of the s level Calculate the energy splitting between these states and relate this to the frequency of the photons SET UP: When the spin component is parallel to the field the interaction energy is U = μzb When the spin component is antiparallel to the field the interaction energy is U =+ μzb The transition energy for a transition between these two states is Δ E = μzb, where μz = 98 μn The transition energy is related to the photon frequency by Δ E = hf, so z 4 hf ( J s)( Hz) B = = = 5 T μz (98)(55 J/T) EVALUATE: This magnetic field is easily achievable Photons of this frequency have wavelength = cf / = m These are radio waves 44 8 (a) As in Example 4, Δ E = (9)(545 ev T)( T) = ev Since μ and S are in opposite directions for a neutron, the antiparallel configuration is lower energy This result is smaller than but comparable to that found in the example for protons ΔE c (b) f = = 9 MHz, = = 448 m h f 45 IDENTIFY: Calculate the spin magnetic energy shift for each spin component Calculate the energy splitting between these states and relate this to the frequency of the photons (a) SET UP: From Example 4, when the z-component of S (and μ ) is parallel to B, U = μ B = 98 μnb When the z-component of S (and μ ) is antiparallel to B, U = μz B =+ 98 μnb The state with the proton spin component parallel to the field lies lower in energy The energy difference between these two states is Δ E = (98 μnb) Δ E (98 μn) B (98)(55 J/T)(5 T) Δ E = hf so f = = = 4 h h J s f = Hz = MHz 8 c 998 m/s And then = = = 4 m f Hz EVALUATE: From Figure 4 in the textbook, these are radio waves z 4-

2 4- Chapter 4 (b) SET UP: From Eqs () and (4) and Fig44 in the textbook, the state with the z-component of μ parallel to B has lower energy But, since the charge of the electron is negative, this is the state with the electron spin component antiparallel to B That is, for the m state s = lies lower in energy For the e e m s =+ U =+ () B () B () + =+ μ =+ BB m m For the ms = state, U = () μ BB The energy difference between these two states is Δ E = () μbb Δ E = hf 4 Δ E μbb ()(94 J/T)(5 T) so f = = = = 4 Hz = 4 GHz And 4 h h J s 8 c 998 m/s = = = 49 m = 49 mm f 4 Hz EVALUATE: From Figure 4 in the textbook, these are microwaves The interaction energy with the magnetic field is inversely proportional to the mass of the particle, so it is less for the proton than for the electron The smaller transition energy for the proton produces a larger wavelength 4 (a) 4mn + 9mH mu = 9 u (b) 8 MeV (c) 5 MeVper nucleon (using 95 MeV/u and 8 nucleons) 4 IDENTIFY and SET UP: The text calculates that the binding energy of the deuteron is 4 MeV A photon that breaks the deuteron up into a proton and a neutron must have at least this much energy hc hc E = so = E 5 8 (4 ev s)(998 m/s) = = 555 m = 555 pm 4 ev EVALUATE: This photon has gamma-ray wavelength 48 IDENTIFY: The binding energy of the nucleus is the energy of its constituent particles minus the energy of the carbon- nucleus SET UP: In terms of the masses of the particles involved, the binding energy is E B = (m H + m n m C- )c (a) Using the values from Table 4, we get E B = [(85 u) + (85 u) u)](95 MeV/u) = 9 MeV (b) The binding energy per nucleon is (9 MeV)/( nucleons) = 8 MeV/nucleon (c) The energy of the C- nucleus is ( u)(95 MeV/u) = 8 MeV Therefore the percent of the mass that is binding energy is 9 MeV 845% 8 MeV = EVALUATE: The binding energy of 9 MeV binds nucleons The binding energy per nucleon, rather than just the total binding energy, is a better indicator of the strength with which a nucleus is bound 49 IDENTIFY: Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the energy after the split (kinetic energy plus energy of the proton and neutron) Therefore the kinetic energy released is equal to the energy of the photon minus the binding energy of the deuteron SET UP: The binding energy of a deuteron is 4 MeV and the energy of the photon is E = hc/ Kinetic energy is K = ½mv (a) The energy of the photon is hc ( 4 J s)( 8 m/s) Eph = = = 58 J 5 m The binding of the deuteron is E B = 4 MeV = 5 J Therefore the kinetic energy is K = (58 5) J = J = MeV (b) The particles share the energy equally, so each gets half Solving the kinetic energy for v gives v = K ( J) = = m/s m 5 kg EVALUATE: Considerable energy has been released, because the particle speeds are in the vicinity of the speed of light 4 (a) ( mn + mh) mn = u, which is 5 MeV, or 48 MeV per nucleon (b) Similarly, ( mh + mn) mhe = 8 u = 8 MeV, or MeV per nucleon, slightly lower (compare to Figure 4 in the textbook)

3 Nuclear Physics 4-4 (a) IDENTIFY: Find the energy equivalent of the mass defect SET UP: A 5B atom has 5 protons, 5 = neutrons, and 5 electrons The mass defect therefore is Δ M = 5mp + mn + 5 me M( 5B) Δ M = 5(5 u) + (849 u) + 5( u) 95 u = 88 u The energy equivalent is E B = (88 u)(95 MeV/u) = MeV / / (b) IDENTIFY and SET UP: Eq(4): EB = CA CA CZ( Z )/ A C4( A Z) / A The fifth term is zero since Z is odd but N is even A= and Z = 5 / / E B = (55 MeV)() (8 MeV)() ( MeV)5(4)/ (9 MeV)( ) / E B =+ 5 MeV 884 MeV 8 MeV 5 MeV = 8 MeV The percentage difference between the calculated and measured E B is 8 MeV MeV = % MeV EVALUATE: Eq(4) has a greater percentage accuracy for Ni The semi-empirical mass formula is more accurate for heavier nuclei 4 (a) 4mn + 9mH mcu = 4(85) u + 9(85) u 99 u = 59 u, which is 55 MeV, or 85 MeV per nucleon (using 95 MeV/u and nucleons) (b) In Eq(4), Z = 9 and N = 4, so the fifth term is zero The predicted binding energy is (9)(8) (5) E B = (55 MeV)() (8 MeV)() ( MeV) (9 MeV) () () E B = 55 MeV The fifth term is zero since the number of neutrons is even while the number of protons is odd, making the pairing term zero This result differs from the binding energy found from the mass deficit by 8%, a very good agreement comparable to that found in Example IDENTIFY In each case determine how the decay changes A and Z of the nucleus The β and β particles have charge but their nucleon number is A = (a) SET UP: α -decay: Z increases by, A= N + Z decreases by 4 (an α particle is a 4 He nucleus) Pu He + 9 U (b) SET UP: β decay: Z increases by, A= N + Z remains the same (a β particle is an electron, e ) 4 4 Na e + Mg (c) SET UP β + decay: Z decreases by, A= N + Z remains the same (a β + particle is a positron, e) O + e+ N EVALUATE: In each case the total charge and total number of nucleons for the decay products equals the charge and number of nucleons for the parent nucleus; these two quantities are conserved in the decay 4 44 (a) The energy released is the energy equivalent of mn mp me = 84 u, or 8 kev (b) m, n > m p and the decay is not possible 45 IDENTIFY: The energy of the photon must be equal to the difference in energy of the two nuclear energy levels SET UP: The energy difference is ΔE = hc/ hc ( 4 J s)( 8 m/s) 5 Δ E = = = 85 J = 5 MeV 9 48 J EVALUATE: Since the wavelength of this photon is much shorter than the wavelengths of visible light, its energy is much greater than visible-light photons which are frequently emitted during electron transitions in atoms This tells us that the energy difference between the nuclear shells is much greater than the energy difference between electron shells in atoms, meaning that nuclear energies are much greater than the energies of orbiting electrons 4 IDENTIFY: The energy released is equal to the mass defect of the initial and final nuclei SET UP: The mass defect is equal to the difference between the initial and final masses of the constituent particles (a) The mass defect is 8588 u 44 u 4 u = 4584 u The energy released is (4584 u)(95 MeV/u) = 4 MeV (b) Take the ratio of the two kinetic energies, using the fact that K = p /m: pth KTh mth mα 4 = = = K p α α mth 4 m α

4 4-4 Chapter 4 The kinetic energy of the Th is 4 4 (4 MeV) = MeV = 48 4 K J Th = KTotal = Solving for v in the kinetic energy gives 4 K (48 J) 5 v = = = 4 m/s m (44) 5 kg ( ) EVALUATE: As we can see by the ratio of kinetic energies in part (b), the alpha particle will have a much higher kinetic energy than the thorium 4 4 If β 4 4 decay of C is possible, then we are considering the decay C N + β 4 4 Δ m= M( C) M( N) me Δ m = (44 u (549 u)) (44 u (549 u)) 549 u 4 4 Δ m=+ 8 u So E = (8 u)(95 MeV u ) = 5 MeV = 5 kev 48 (a) A proton changes to a neutron, so the emitted particle is a positron ( β + ) (b) The number of nucleons in the nucleus decreases by 4 and the number of protons by, so the emitted particle is an alpha-particle (c) A neutron changes to a proton, so the emitted particle is an electron ( β ) 49 (a) As in the example, (898 u)(95 MeV u) = 8 MeV (b) 8 MeV MeV 4 MeV = MeV (a) Sr β + X X has 9 protons and 9 protons plus neutrons, so it must be 9 Y 9 9 (b) Use base because we know the half life A= A and A = A tt T log (8 yr)log t = = = 9yr log log ln 5 4 IDENTIFY and SET UP: T/ = The mass of a single nucleus is 4mp = kg ΔN/ Δ t = 5 Ci = Bq ; ΔN/ Δ t = N kg ΔN/ Δ t Bq N = = 9 ; = = = 49 s 5 kg N 9 ln T 4 / = = 58 s = 5 yr / 4 Note that Eq(4) can be written as follows: N = N t T The amount of elapsed time since the source was (5 yr)/(5 yr) created is roughly 5 years Thus, we expect the current activity to be N = (5 Ci) = Ci The ln() source is barely usable Alternatively, we could calculate = = (years) and use the Eq 4 directly T to obtain the same answer 4 IDENTIFY and SET UP: As discussed in Section 44, the activity A= dn/ obeys the same decay equation as 4 (ln ) t/ T/ Eq (4): A= Ae For C, T = 5 y and = ln/ T so A= Ae ; Calculate A at each t; tt / / A = 8 decays/min (a) t = y, A= 59 decays/min (b) t = 5, y, A= 4 decays/min 8 EVALUATE: The time in part (b) is 8 half-lives, so the decay rate has decreased by a factor or ( ) 44 IDENTIFY and SET UP: The decay rate decreases by a factor of in a time of one half-life (a) 4 d is T / so the activity is (5 Bq)/( ) = 49 Bq (b) The activity is proportional to the number of radioactive nuclei, so the percent is Bq % 49 Bq = (c) 5 I e+ 54Xe The nucleus 54 Xe is produced EVALUATE: Both the activity and the number of radioactive nuclei present decrease by a factor of in one halflife

5 45 (a) H e+ He (b) N = N e, N = N and = (ln ) T (ln ) = e t T ; t(ln ) T = ln(); dn 4 (a) = = = ln ln ln T = = = = 9 s T d(8,4s d) 5 μci (5 )( s ) 85 decays s t ln() T = = ln Nuclear Physics y dn dn 85 decays / s = N N = = = nuclei The mass of this many Ba nuclei is 9 s 9 m = nuclei ( kg nucleus ) = kg = g = ng (b) A= Ae μci = (5 μci) e ln(/5) = t ln( 5) ln( 5) d t = = = = t(ln )/ T (ln ) t / 4 A= Ae = Ae = ln( A A ) T s 8 days 9 s 8,4 s (ln ) t (ln )(4 days) T = = = 8 days ln( A A ) ln(9 88) dn ln ln = N = = = s T yr 5 s/yr N = = g ( ) atoms 5 g 5 atoms dn = N = = = Ci Convert to Ci: Bq 98 Ci = Bq 5 (5 )( s ) decays/s Bq 49 IDENTIFY and SET UP: Calculate the number N of 4 C atoms in the sample and then use Eq (4) to find the decay constant Eq (48) then gives / Find the total number of carbon atoms in the sample n= m/ M; N nn mn tot A A / = = = ( kg)( atoms/mol)/( kg/mol) Ntot = atoms, so ( )( ) = 8 carbon-4 atoms ΔN/ Δ t = 8 decays/min = decays/s ΔN/ Δ t = N; ΔN/ Δt = = 8 s N T/ = (ln )/ = 8 s = 5 y EVALUATE: The value we calculated agrees with the value given in Section 44 4 decays = 4 Bq = Ci = μci 8,4 s 4 dn (a) = 5 Bq = 5 decays s = = = 5 s T (8 min)( s min) dn 5 decays s N = = = 4 5 s 5 nuclei

6 4- Chapter 4 (b) The number of nuclei left after one half-life is dn 8 decays s = N = 5 nuclei, and the activity is half: 4 (c) After three half lives (94 minutes) there is an eighth of the original amount, so N = 5 nuclei, and an dn eighth of the activity: = 945 decays s decays min 4 The activity of the sample is = Bq kg, while the activity of atmospheric carbon is ( sec min) (5 kg) ln ( 55) ln ( 55) 55 Bq kg (see Example 49) The age of the sample is then t = = = 5 y 4 y 4 IDENTIFY and SET UP: Find from the half-life and the number N of nuclei from the mass of one nucleus and the mass of the sample Then use Eq(4) to calculate dn /, the number of decays per second (a) dn / = N 9 9 = = = 5 s 9 T/ (8 y)(5 s/ y) The mass of 4 K atom is approximately 4 u, so the number of 4 K nuclei in the sample is 9 9 kg kg N = = = u 4(54 kg) Then dn / = N = (5 s )(454 ) = 4 decays/s (b) dn / = (4 decays/s)( Ci/( decays/s)) = 4 Ci EVALUATE: The very small sample still contains a very large number of nuclei But the half life is very large, so the decay rate is small 44 (a) rem = rad RBE = x() and x = rad (b) rad deposits J kg, so rad deposit J kg This radiation affects 5 g (5 kg) of tissue, so the total energy is (5 kg)( J kg) = 5 J = 5 mj (c) Since RBE = for β -rays, so rem = rad Therefore rad = rem 45 rad = Gy, so Gy = rad and the dose was 5 rad rem = (rad)(rbe) = (5 rad)(4) = rem Gy = J kg, so 5 J kg 4 IDENTIFY and SET UP: For x rays RBE = so the equivalent dose in Sv is the same as the absorbed dose in J/kg One whole-body scan delivers (5 kg)( J/kg) = 9 J One chest x ray delivers 9 J (5 kg)( J/kg) = J It takes = 9 chest x rays to deliver the same total energy J 4 IDENTIFY and SET UP: For x rays RBE = and the equivalent dose equals the absorbed dose (a) 5 krad = 5 krem = 5 kgy = 5 ksv (5 J/kg)(5 kg) = J (b) 5 krad = 5 kgy ; (5)(5 krad) = krem = ksv The energy deposited would be J, the same as in (a) EVALUATE: The energy required to raise the temperature of 5 kg of water C is 8 J, and J is less than this The energy deposited corresponds to a very small amount of heating 48 (a) 54 Sv ( rem Sv) = 54 rem (b) The RBE of gives an absorbed dose of 54 rad (c) The absorbed dose is 54 Gy, so the total energy absorbed is (54 Gy) (5 kg) = 5 J The energy required to raise the temperature of 5 kg by C is (5 kg) (49 J kg K) (C ) = kj 49 (a) We need to know how many decays per second occur = = = 9 s T ( y) (5 s y) The number of tritium atoms is dn (5 Ci) ( Bq Ci) 9 N = = = 9 s 8 54 nuclei

7 44 Nuclear Physics 4- The number of remaining nuclei after one week is 9 8 (9 s ) () (4) (s) 8 5 N = N e = (5 ) e = 4 nuclei Δ N = N N = 8 decays So the 5 9 energy absorbed is Etotal =Δ N E γ = (8 ) (5 ev) ( J ev) = 4 J The absorbed dose is (4 J) = 5 J kg = 5 rad Since RBE =, then the equivalent dose is 5 rem (5 kg) (b) In the decay, antinetrinos are also emitted These are not absorbed by the body, and so some of the energy of the decay is lost (about kev ) ( Ci) ( Bq Ci) (5 s) = 84 α particles The absorbed dose is 9 (84 ) (4 ev) ( J ev) = 8 Gy = 8 rad The equivalent dose is () (8 rad) = rem (5 kg) 44 (a) IDENTIFY and SET UP: Determine X by balancing the charge and nucleon number on the two sides of the reaction equation X must have A= + 4 = and Z = + 5 = Thus X is Li and the reaction is 4 H+ N Li+ 5B (b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent The neutral atoms on each side of the reaction equation have a total of 8 electrons, so the electron masses cancel when neutral atom masses are used The neutral atom masses are found in Table 4 4 mass of H + N is 4 u + 44 u = u mass of Li + 5B is 5 u + 9 u = 858 u The mass increases, so energy is absorbed by the reaction The Q value is ( u 858 u)(95 MeV/u) = 4 MeV (c) IDENTIFY and SET UP: The available energy in the collision, the kinetic energy Kcm in the center of mass reference frame, is related to the kinetic energy K of the bombarding particle by Eq (44) The kinetic energy that must be available to cause the reaction is 4 MeV Thus K = 4 MeV The mass M of the stationary target ( N) is M = 4 u The mass m of the colliding particle cm ( H)is u Then by Eq (44) the minimum kinetic energy K that the H must have is M m 4 u u K + K + = cm = (4 MeV) = 59 MeV M 4 u EVALUATE: The projectile ( H ) is much lighter than the target 4 ( ) 4 N so K is not much larger than K The K cm we have calculated is what is required to allow the mass increase We would also need to check to see if at this energy the projectile can overcome the Coulomb repulsion to get sufficiently close to the target nucleus for the reaction to occur 44 m + m m4 m = 9 u, so the energy released is 84 MeV He H He H 44 IDENTIFY and SET UP: Determine X by balancing the charge and the nucleon number on the two sides of the reaction equation X must have A = = and Z=+ + 4 = Thus X is Li and the reaction is 9 4 H+ 4Be= Li+ He (b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent If we use the neutral atom masses then there are the same number of electrons (five) in the reactants as in the products Their masses cancel, so we get the same mass defect whether we use nuclear masses or neutral atom masses The neutral atoms masses are given in Table 4 9 H + 4Be has mass 4 u + 98 u = 84 u 4 Li + He has mass u + 4 u = 8 u The mass decrease is 84 u 8 u = 8 u This corresponds to an energy release of 8 u(95 MeV/ u) = 5 MeV (c) IDENTIFY and SET UP: Estimate the threshold energy by calculating the Coulomb potential energy when the 9 H and 4Be nuclei just touch Obtain the nuclear radii from Eq (4) 9 5 / 5 The radius RBe of the 4Be nucleus is RBe = ( m)(9) = 5 m 5 / 5 The radius RH of the H nucleus is RH = ( m)() = 5 m The nuclei touch when their center-to-center separation is 5 R= RBe + RH = 4 m

8 4-8 Chapter 4 The Coulomb potential energy of the two reactant nuclei at this separation is qq e(4 e) U = = 4πP r 4πP r 9 9 4( C) U = (8988 N m /C ) = 4 MeV 5 9 (4 m)( J/eV) This is an estimate of the threshold energy for this reaction EVALUATE: The reaction releases energy but the total initial kinetic energy of the reactants must be 4 MeV in order for the reacting nuclei to get close enough to each other for the reaction to occur The nuclear force is strong but is very short-range 444 IDENTIFY and SET UP: % of naturally occurring uranium is the isotope 5 U The mass of one 5 U nucleus is about 5m p 9 J 9 (a) The number of fissions needed is = The mass of 9 ( ev)( J/eV) U required is ( )(5 m ) = kg p 5 kg (b) = kg EVALUATE: The calculation assumes % conversion of fission energy to electrical energy 445 IDENTIFY and SET UP: The energy released is the energy equivalent of the mass decrease u is equivalent to 95 MeV The mass of one 5 U nucleus is 5m p (a) 9 U+ n 5Ba+ Kr+ n We can use atomic masses since the same number of electrons are included on each side of the reaction equation and the electron masses cancel The mass decrease is Δ M = m( 5 ) ( ) ( 44 ) ( 89 ) ( 9 U + m n m 5Ba + m Kr + m n) Δ M = 549 u u 4995 u 889 u (849 u) Δ M = 8 u The energy released is (8 u)(95 MeV/u) = MeV kg (b) The number of 5 U nuclei in g is = 55 The energy released per gram is 5m ( MeV/nucleus)(55 nuclei/g) = 44 MeV/g 8 4 A 44 (a) Si + γ Mg + Z X A+ 4 = 8 so A= 4 Z + = 4 so Z = X is an α particle 4 Eγ (b) = Δ mc = (9854 u + 4 u 99 u) (95 MeV u ) = 9984 MeV 44 The energy liberated will be 4 M( He) + M( He) M( 4Be) = (9 u + 4 u 99 u)(95 MeV u) = 58 MeV 448 (a) Z = + = 5and A= 4+ = (b) The nuclide is a boron nucleus, and mhe + mli mn mb = u, and so 9 MeV of energy is absorbed A Z+ A 4 ( Z ) Nuclei: ZX Z Y + He Add the mass of Z electrons to each side and we find: A A 4 4 Δ m= M( ZX) M( Z Y) M( He), where now we have the mass of the neutral atoms So as long as the mass of the original neutral atom is greater than the sum of the neutral products masses, the decay can happen A A 45 Denote the reaction as ZX Z+ Y+ e The mass defect is related to the change in the neutral atomic masses by [ mx Zme] [ my ( Z + ) me] me = ( mx my), where mx and m Y are the masses as tabulated in, for instance, Table (4) ( ) 45 A X Z + A Y Z + + A + A + Z Z + β Adding (Z ) electrons to both sides yields ZX Z Y+ β So in terms of masses: A + A A A A A ( Z ) ( Z ) ( ( Z ) ) ( Z ) ( Z ) ( Z ) Δ m= M X M Y me = M X me M Y me = M X M Y me So the decay will occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses 45 IDENTIFY and SET UP: m= ρv gal = 88 L = 88 m The mass of a 5 U nucleus is 5m p MeV = J (a) For gallon, m= ρv = ( kg/m )(88 m ) = 9 kg = 9 g 8 J/gal 4 = 4 J/g 9 g/gal p

9 kg (b) g contains = 55 nuclei 5m p ( MeV/nucleus)( J/MeV)(55 nuclei) = 8 J/g (c) A mass of m p produces MeV ( MeV)( J/MeV) 4 = 4 J/kg = 4 J/g m p (d) The total energy available would be (99 kg)(4 J/kg) = 94 J energy energy 94 J power = so t = = = 4 s = yr t power 8 W EVALUATE: If the mass of the sun were all proton fuel, it would contain enough fuel to last 4 J/g ( yr) yr 4 = 4 J/g A 4 45 Using Eq: (4): Z M = ZM + H Nm n EB c M ( Na) = M + H m n EB/ c ()() But E B = (55 MeV)(4) (8 MeV)(4) ( MeV) (4) (4 ()) 4 (9 MeV) (9 MeV)(4) = 98 MeV 4 4 (98 MeV) M ( Na) = (85 u) + ( 85 u) = 9858 u 95 MeV u % error = = % If the binding energy term is neglected, M ( Na) = 498u and the percentage error would be = 8% The α -particle will have of the released energy (see Example 45) ( Th Ra ) m m m α = 5 u or 49 MeV Nuclear Physics (a) IDENTIFY and SET UP: The heavier nucleus will decay into the lighter one 5 5 Al will decay into Mg (b) IDENTIFY and SET UP: Determine the emitted particle by balancing A and Z in the decay reaction 5 5 This gives Al Mg + + e The emitted particle must have charge + e and its nucleon number must be zero Therefore, it is a β + particle, a positron (c) IDENTIFY and SET UP: Calculate the energy defect Δ M for the reaction and find the energy equivalent of Δ M Use the nuclear masses for Al and Mg, to avoid confusion in including the correct number of electrons if 5 5 neutral atom masses are used 5 5 The nuclear mass for Al is M nuc ( Al) = u (54858 u) = 4989 u 5 5 The nuclear mass for Mg is M nuc ( Mg) = u (54858 u) = u The mass defect for the reaction is 5 5 Δ M = Mnuc ( Al) Mnuc ( Mg) M ( + e) = 4989 u u u = 494 u Q= ( Δ M) c = 494 u(95 MeV/ u) = 55 MeV EVALUATE: The mass decreases in the decay and energy is released Note: 5 Al can also decay into 5 Mg by the electron capture 5 5 Al + e Mg 5 The electron in the reaction is an orbital electron in the neutral Al atom The mass defect can be calculated using the nuclear masses: 5 5 Δ M = M Al + M ( e) M Mg = 4988 u u u = 459 u ( ) ( ) nuc nuc ( ) M Q= Δ c = (459 u)(95 MeV/ u) = 4 MeV The mass decreases in the decay and energy is released

10 4- Chapter 4 45 (a) m m m4 = 58 u, or Q= 54 MeV The energy of the alpha particle is ( ) times this, 84 Po 8 Pb He or 5 MeV (see Example 45) (b) m m9 m = 55 u <, so the decay is not possible Po Bi H 84 8 (c) m m9 mn 8 u, 84 Po 84 Po (d) m m, 85 At 84 Po = < so the decay is not possible > so the decay is not possible (see Problem (45)) m + m > m, so the decay is not possible (see Problem (45)) (e) 8 Bi e 84 Po 45 IDENTIFY and SET UP: The amount of kinetic energy released is the energy equivalent of the mass change in the decay m e = 548 u and the atomic mass of 4 N is 44 u The energy equivalent of u is 95 MeV 4 C has a half-life of T / = 5 yr = 8 s The RBE for an electron is 4 4 C e + N+ υ (a) e Δ M = m 4 4 C me + m Use nuclear masses, to avoid difficulty in accounting for atomic electrons The nuclear mass of 4 C is 44 u me = u The nuclear mass of 4 N is 44 u m = 9994 u (b) The mass decrease is ( ) ( N) e 4 Δ M = u 9994 u 549 u = u The energy equivalent of Δ M is 5 MeV (c) The mass of carbon is (8)(5 kg) = 5 kg From Example 49, the activity due to g of carbon in a living organism is 55 Bq The number of decay/s due to 5 kg of carbon is (5 )(55 Bq/g) = 4 decays/s (d) Each decay releases 5 MeV so 4 decays/s releases 5 MeV/s 85 J/s = (e) The total energy absorbed in yr is (85 J/s)(5 s) = J The absorbed dose is J 5 = J/kg = μgy = mrad With RBE =, the equivalent dose is μ Sv = mrem 5 kg IDENTIFY and SET UP: mπ = 4me = 4 kg The total energy of the two photons equals the rest mass energy m π c of the pion 8 8 (a) E m c ph = (4 kg)( m/s) 8 J 5 MeV π = = = hc hc 4 ev m 4 Eph = so 84 = = = m = 84 fm Eph 5 ev These are gamma ray photons, so they have RBE = (b) Each pion delivers (8 J) = J The absorbed dose is rad = Gy = J/kg The energy deposited is (5 kg)( J/kg) = 5 J 5 J 9 The number of π mesons needed is = mesons J/meson EVALUATE: Note that charge is conserved in the decay since the pion is neutral If the pion is initially at rest the photons must have equal momenta in opposite directions so the two photons have the same and are emitted in opposite directions The photons also have equal energies since they have the same momentum and E = pc 459 IDENTIFY and SET UP: Find the energy equivalent of the mass decrease Part of the released energy appears as the emitted photon and the rest as kinetic energy of the electron Au 8 Hg + e The mass change is 9985 u 995 u = 4 u (The neutral atom masses include 9 electrons before the decay and 8 electrons after the decay This one additional electron in the products accounts correctly for the electron emitted by the nucleus) The total energy released in the decay is (4 u)(95 MeV/u) = MeV This energy is divided between the energy of the emitted photon and the kinetic energy of the β particle Thus the β particle has kinetic energy equal to MeV 4 MeV = 9 MeV EVALUATE: The emitted electron is much lighter than the 98 8 Hg nucleus, so the electron has almost all the final kinetic energy The final kinetic energy of the 98 Hg nucleus is very small

11 Nuclear Physics 4-4 (See Problem (45)) m m me = u Decay is energetically possible C 5B 4 IDENTIFY and SET UP: The decay is energetically possible if the total mass decreases Determine the nucleus produced by the decay by balancing A and Z on both sides of the equation N + e+ C To avoid confusion in including the correct number of electrons with neutral atom masses, use nuclear masses, obtained by subtracting the mass of the atomic electrons from the neutral atom masses N is M N = 59 u (54858 u) = 899 u The nuclear mass for ( ) nuc The nuclear mass for nuc ( ) C is M C = 55 u (54858 u) = 4 u The mass defect for the reaction is Δ M = M N M C M e Δ M = 899 u 4 u u = 8 u EVALUATE: ( ) ( ) ( + ) nuc nuc The mass decreases in the decay, so energy is released This decay is energetically possible 4 (a) A least-squares fit to log of the activity vs time gives a slope of = 5995 h, for a half-life of ln h = (b) The initial activity is N, and this gives 4 ( Bq) N = = (5995 hr )( hr s) t (c) Ne = 8 dn() t dn() t 4 The activity At ( ) but = N() t so N = A Taking the derivative of dn() t t Nt () = Ne = Ne = Ae, or At ( ) = Ae t t (ln ) ( t/ T) 44 From Eq4 Nt () Ne = butne = Ne ( / ) ( / ) (ln ) ln(/ ) t = N e = N e So N( t) = N where n= T a ax x a ln x (We have used that aln x = ln( x ), e = ( e ), and e = x ) 45 IDENTIFY and SET UP: One-half of the sample decays in a time of T / 9 yr 4 (a) = 5, yr t T t T n 8 4 (b) 5 ( ) This exponent is too large for most hand-held calculators But ( ) = so () = ( ) = , ln 5 4 IDENTIFY and SET UP: T/ = The mass of a single nucleus is 49mp = 49 kg ΔN/ Δ t = N kg N = = 48 ΔN/ Δ t = 5 decays/s 5 49 kg ΔN/ Δt 5 decays/s ln = = = 55 s ; T 4 / = = s = 99 yr N 48 4 IDENTIFY: Use Eq (4) to relate the initial number of radioactive nuclei, N, to the number, N, left after time t SET UP: We have to be careful; after 8 Rb has undergone radioactive decay it is no longer a rubidium atom Let N 85 be the number of 85 Rb atoms; this number doesn t change Let N be the number of 8 Rb atoms on earth when the solar system was formed Let N be the present number of 8 Rb atoms The present measurements say that 8 = N/( N + N85) ( N + N85)(8) = N, so N = 85 N85 The percentage we are asked to calculate is N/( N + N85) t + t N and N are related by N = Ne so N = e N t t t N Ne (855 e ) N85 85e Thus = = = t t t N + N85 Ne + N85 (85 e ) N85 + N85 85e t = 4 y; = = = 459 y T 45 y / e e e 9 t (459 y )(4 y) = = = 94

12 4- Chapter 4 N (85)(94) Thus 9% N + N = 85 (85)(94) + = EVALUATE: The half-life for 8 Rb is a factor of larger than the age of the solar system, so only a small fraction of the 8 Rb nuclei initially present have decayed; the percentage of rubidium atoms that are radioactive is only a bit less now than it was when the solar system was formed 9 48 (a) (5 )(4 MeV)( J ev) ( kg) = 8 Gy = 8 rad (b) ()(8 rad ) = rem m ln() 9 (c) N = Bq mci Amp T = = 5 (d) = 54 s, about an hour and a half Note that this time is so small in comparison with the 9 Bq half-life that the decrease in activity of the source may be neglected 49 IDENTIFY and SET UP: Find the energy emitted and the energy absorbed each second Convert the absorbed energy to absorbed dose and to equivalent dose (a) First find the number of decays each second: 4 decays/s Ci = 9 decays/s Ci The average energy per decay is 5 MeV, and one-half of this energy is deposited in the tumor The energy delivered to the tumor per second then is 9 (9 decays/s)(5 ev/decay)( J/eV) = 9 J/s (b) The absorbed dose is the energy absorbed divided by the mass of the tissue: 9 J/s 4 = (9 J/kg s)( rad/( J/kg)) = 9 rad/s 5 kg (c) equivalent dose (REM) = RBE absorbed dose (rad) 4 4 In one second the equivalent dose is (9 rad) = rem 4 (d) ( rem/ rem/s) = /5 s( h/ s) = 4 h = days EVALUATE: The activity of the source is small so that absorbed energy per second is small and it takes several days for an equivalent dose of rem to be absorbed by the tumor A rem dose equals Sv and this is large enough to damage the tissue of the tumor 4 (4) 9 4 (a) After 4 min = 4 s, the ratio of the number of nuclei is = = (b) After 5 min = 9 s, the ratio is 5 4 IDENTIFY and SET UP: The number of radioactive nuclei left after time t is given by N = N e The problem says N/ N = ; solve for t = e so ln() = t and t = ln()/ 4 4 ln() 4 Example 49 gives = 9 y for C Thus t = = y 4 9 y EVALUATE: remaining is less than ( ) The half-life of 4 C is 5 y, so our calculated t is more than two half-lives, so the fraction = 4 4 IDENTIFY: The tritium (H-) decays to He- The ratio of the number of He- atoms to H- atoms allows us to calculate the time since the decay began, which is when the H- was formed by the nuclear explosion The H- decay is exponential SET UP: The number of tritium (H-) nuclei decreases exponentially as NH = N,He, with a half-life of years The amount of He- present after a time t is equal to the original amount of tritium minus the number of tritium nuclei that are still undecayed after time t The number of He- nuclei after time t is N = N N = N N e = N e He,H H,H,H,H ( ) Taking the ratio of the number of He- atoms to the number of tritium (H-) atoms gives,h ( t N N e ) He e t = = = e N N e e H,H

13 ( ) Nuclear Physics 4- ln + NHe / NH ln Solving for t gives t = Using the given numbers and T/ =, we have = ln ln 5/ y T = / y = and ln( + 4) t = = years 5/ y EVALUATE: One limitation on this method would be that after many years the ratio of H to He would be too small to measure accurately 4 (a) IDENTIFY and SET UP: Use Eq(4) to calculate the radius R of a H nucleus Calculate the Coulomb potential energy (Eq9) of the two nuclei when they just touch The radius of H is 5 / 5 R = ( m)() = 5 m The barrier energy is the Coulomb potential energy of two H nuclei with their centers separated by twice this distance: 9 e 9 ( C) 4 U = = (8988 N m / C ) = 4 J = 48 MeV 5 4π P r (5 m) (b) IDENTIFY and SET UP: Find the energy equivalent of the mass decrease H+ H He+ n If we use neutral atom masses there are two electrons on each side of the reaction equation, so their masses cancel The neutral atom masses are given in Table 4 H + H has mass (4 u) = 484 u He + n has mass 9 u + 85 u = 4494 u The mass decrease is 484 u 4494 u = 5 u This corresponds to a liberated energy of 9 (5 u)(95 MeV/u) = MeV, or ( ev)( J/eV) = 59 J (c) IDENTIFY and SET UP: We know the energy released when two H nuclei fuse Find the number of reactions obtained with one mole of H Each reaction takes two H nuclei Each mole of D has molecules, so pairs of atoms The energy liberated when one mole of deuterium undergoes fusion is ( )(59 J) = 55 J/mol EVALUATE: The energy liberated per mole is more than a million times larger than from chemical combustion of one mole of hydrogen gas 44 In terms of the number N of cesium atoms that decay in one week and the mass m = kg, the equivalent dose is N N N 5 Sv = ((RBE) γe γ + (RBE) ee e) = (()( MeV) + (5)(5 MeV)) = (8 J), so m m m ( kg)(5 Sv) N = = 55 The number N of atoms present is related to (8 J) t ln 9 N by N = Ne = = sec Ty ( yr)(5 sec yr) 4 t ( s )( days)(84 s day) Then N = Ne = (55 ) e = 5 45 (a) v v m m M vm cm = v m + M m = v v = v v M = m+ M m+ M m+ M mm Mm M mm m K = mv m + Mv M = v + v = v + ( m+ M) ( m+ M) ( m+ M) m+ M m+ M M M K = mv K = K Kcm m+ M m+ M (b) For an endoergic reaction Kcm Q( Q ) M ( M + m) Q= Kth Kth = Q M + m M = < at threshold Putting this into part (a) gives

14 4-4 Chapter 4 4 Mα K = K, where K is the energy that the α -particle would have if the nucleus were infinitely massive Mα + m 8 Then, M = MOs Mα K = MOs Mα ( MeV c ) = 8948u 8 Δ m= M U M 54 Xe M 8Sr mn Δ m = 549 u 99 u 995 u 85 u = 98 u E = Δ m c = 98 u 95 MeV u = 85 MeV 4 ( ) ( ) ( ) ( ) ( )( ) 48 (a) A least-squares fit of the log of the activity vs time for the times later than 4 h gives a fit with correlation ( ) and decay constant of h, corresponding to a half-life of 9 h Extrapolating this back to time gives a contribution to the rate of about 5/s for this longer-lived species A least-squares fit of the log of the activity vs time for times earlier than h gives a fit with correlation = 994, indicating the presence of only two species ( h) ( h) (b) By trial and error, the data is fit by a decay rate modeled by ( 5 Bq) t t R= e + ( 5 Bq) e This would correspond to half-lives of 4 h and 9 h (c) In this model, there are 4 of the shorter-lived species and 49 of the longer-lived species (d) After 5 h, there would be 8 of the shorter-lived species and 4 of the longer-lived species 49 (a) There are two processes occurring: the creation of 8 I by the neutron irradiation, and the decay of the newly produced 8 I So dn = K N where K is the rate of production by the neutron irradiation Then t N dn t N K( e ) = K N ln( K N ) = ln( K N) = ln K t N() t = The graph is given in Figure 49 9 t 5 min (b) The activity of the sample is N() t = K( e ) = ( 5 decays s) e So the activity is t ( 5 decays s)( e dn ), with t in minutes So the activity at various times is: (c) N max dn dn t = = t = = dn dn t = = t = = dn dn t = = t = = K (5 ) () 9 = = = atoms ( ) 4 5 ( min) 4 Bq; ( min) Bq; 5 ( 5 min) 5 Bq; ( 5 min) Bq; ( 5 min) Bq; ( 8 min) 5 Bq; (d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals 5 decays s Figure The activity of the original iron, after hours of operation, would be ( h) (45 d 4 h d) 5 (94 Ci) ( Bq Ci) = 8 Bq The activity of the oil is 84 Bq, or of the total iron activity, and this must be the fraction of the mass worn, or mass of 459 g 5 The rate at which the piston rings lost their mass is then 459 g h