Chapter 45 Solutions

Transcription

1 Chapter 45 Solutions *45.1 m (m n + M U ) (M Zr + M Te + 3m n ) m ( u u) ( u u + 3( u)) m u kg so Q mc J 19 MeV 45. Three different fission reactions are possible: 1 0n + 9U 38Sr + 54Xe + 0n Xe n + 9U 38Sr + 54Xe n Xe n + 9U 38Sr + 54Xe n Xe n Th Th Pa + e + ν 33 91Pa 33 9 U + e + ν n U 39 9 U Np + e + ν 39 93Np Pu + e + ν 45.5 (a) Q ( m)c [ m n + M U35 M Ba141 M Kr9 3m n ]c m [( ) ( ) ]u u Q ( u) ( MeV u) 01 MeV f m u m i u % 45.6 If the electrical power output of 1000 MW is 40.0% of the power derived from fission reactions, the power output of the fission process is 1000 MW J s d J 1 fission 1 ev The number of fissions per day is d ev J 14 J/d This also is the number of 35 U nuclei used, so the mass of 35 U used per day is 4 nuclei 35 g/mol g/d.63 kg/d d nuclei/mol In contrast, a coal-burning steam plant producing the same electrical power uses more than kg/d of coal. d 000 by Harcourt, Inc. All rights reserved.

2 Chapter 45 Solutions The available energy to do work is 0.00 times the energy content of the fuel U 1000 g 1 mol ( 08)( ( 100. kg fuel) fuel 1 kg 35 g mol fission ( J) ( 0.00) J ( N) d d m 5.80 Mm J) Goal Solution Suppose enriched uranium containing 3.40% of the fissionable isotope 35 9 U is used as fuel for a ship. The water exerts an average frictional drag of N on the ship. How far can the ship travel per kilogram of fuel? Assume that the energy released per fission event is 08 MeV and that the ship's engine has an efficiency of 0.0%. G : O : Nuclear fission is much more efficient for converting mass to energy than burning fossil fuels. However, without knowing the rate of diesel fuel consumption for a comparable ship, it is difficult to estimate the nuclear fuel rate. It seems plausible that a ship could cross the Atlantic ocean with only a few kilograms of nuclear fuel, so a reasonable range of uranium fuel consumption might be 10 km / kg to km/kg. The fuel consumption rate can be found from the energy released by the nuclear fuel and the work required to push the ship through the water. A : One kg of enriched uranium contains 3.40% 35 9 U so m 35 ( 1000 g) ( ) 34.0 g In terms of number of nuclei, this is equivalent to 1 N 35 ( 34.0 g) 35 g / mol ( atoms / mol) nuclei If all these nuclei fission, the thermal energy released is equal to ( nuclei) 08 MeV JeV nucleus J Now, for the engine, efficiency work output heat input or e fdcosθ Q h So the distance the ship can travel per kilogram of uranium fuel is d eq h f cos J N m L : The ship can travel km/kg of uranium fuel, which is on the high end of our prediction range. The distance between New York and Paris is km, so this ship could cross the Atlantic ocean on just one kilogram of uranium fuel. 000 by Harcourt, Inc. All rights reserved.

3 594 Chapter 45 Solutions 45.8 (a) For a sphere: V 4 3 πr 3 and r 3V 4π 13 so A V 4πr 1/3 4.84V 3 (4/3)πr For a cube: V l 3 and l V 13 so (c) For a parallelepiped: V a 3 and a V 1/3 so A V 6l 1/3 6V 3 l A V a + 8a a V 1/3 (d) Therefore, the sphere has the least leakage and the parallelepiped has the greatest leakage for a given volume mass of 35 U available ( 0.007) 10 9 metric tons 10 6 g 1 metric ton g number of nuclei ~ g nuclei 35 g mol mol nuclei The energy available from fission (at 08 MeV/event) is E ~ ( events) 08 MeV / event J J / MeV This would last for a time of t E P ~ J Js s 1 yr ~ 3000 yr s In one minute there are 60.0 s 1.0 ms fissions. So the rate increases by a factor of ( ) P 10.0 MW J/s If each decay delivers 1.00 MeV J, then the number of decays/s Bq

4 Chapter 45 Solutions (a) The Q value for the D-T reaction is MeV. Heat content in fuel for D-T reaction: 13 ( MeV )( J/MeV) ( 5 u )( kg/u) 14 J/kg r DT ( J / s)(3600 s / hr) ( J / kg)( g/h burning of D and T kg / g) Heat content in fuel for D-D reaction: Q 1 ( ) 3.65 MeV average of two Q values 13 ( 3.65 MeV) ( J/MeV) ( 4 u )( kg/u) 13 J/kg r DD ( J / s)(3600 s / hr) ( J / kg)( g/h burning of D kg / g) (a) At closest approach, the electrostatic potential energy equals the total energy E. ( Z e) U f k e Z 1 e E: E r min ( N m C )( C) Z 1 Z ( J)Z 1 Z m For both the D-D and the D-T reactions, Z 1 Z 1. Thus, the minimum energy required in both cases is E MeV J MeV J (a) r f r D + r T m U f k e e r f [ 13 + ( 3) 13 ] ( N m C )( C) m m J 444 kev (c) m Conserving momentum, m D v i ( m D + m T )v f, or v f D v i m D + m T 5 v i (d) K i + U i K f + U f : K i ( m D + m T)v f + U f 1 m D + m T m D m D + m T m K i + 0 D m D + m T ( 1 m D v i m )+ U f D K i + U f m D + m T m 1 D m K i U f : K i U D + m T f kev m D + m T m T kev v i + U f (e) Possibly by tunneling. 000 by Harcourt, Inc. All rights reserved.

5 596 Chapter 45 Solutions (a) Average KE per particle is 3 k B T 1 mv. Therefore, v rms 3k B T m ( K) J/K kg m/s t x v ~ 0.1 m 10 6 m/s ~ 10 7 s 1609 m (a) V mi 3 1 mi m 3 m water ρv ( 10 3 kg m 3 )( m 3 ) kg m H M H M H O m H O ( kg) kg m Deuterium ( % )m H ( )( kg) kg The number of deuterium nuclei in this mass is N m Deuterium m Deuteron kg.014 u kg u Since two deuterium nuclei are used per fusion, 1 H + 1 H 4 He + Q, the number of events is N The energy released per event is [ ] c Q M H + M H M 4 He The total energy available is then E N Q MeV [ ]u ( MeV u) 3.8 MeV J 1 MeV J The time this energy could possibly meet world requirements is ( s) t E P J Js 1 yr s yr ~ 1 billion years.

6 Chapter 45 Solutions (a) Including both ions and electrons, the number of particles in the plasma is N nv where n is the ion density and V is the volume of the container. Application of Equation 1.6 gives the total energy as E 3 Nk B T 3nVk B T cm 3 E J ( 50 m 3 ) 106 cm 3 1 m 3 ( K) JK From Table 0., the heat of vaporization of water is L v Jkg. The mass of water that could be boiled away is m E J L v kg Jkg (a) Lawson s criterion for the D-T reaction is nτ scm 3. For a confinement time of τ 1.00 s, this requires a minimum ion density of n cm 3 (c) At the ignition temperature of T K and the ion density found above, the plasma pressure is 106 cm 3 P nk B T cm 3 1 m 3 ( K) JK The required magnetic energy density is then. J m 3 u B B 10 P J m 3 µ J m 3, B 4π ( 10 7 NA )( J m 3 ) 1.77 T Let the number of 6 Li atoms, each having mass u, be N 6 while the number of 7 Li atoms, each with mass u, is N 7. Then, N % of N total ( N 6 + N 7 ), or N Also, total mass [ N 6 ( u)+ N 7 ( u) ]( kg u).00 kg, or N 6 ( u) ( kg u).00 kg. This yields N as the number of 6 Li atoms and N as the number of 7 Li atoms. 000 by Harcourt, Inc. All rights reserved.

7 598 Chapter 45 Solutions 45.0 The number of nuclei in 1.00 metric ton of trash is N 1000 kg( 1000 g / kg) ( nuclei / mol) ( 56.0 g / mol) nuclei At an average charge of 6.0 e/nucleus, q ( )( 6.0) ( ) C Therefore t q I s 1.4 h 45.1 N 0 mass present mass of nucleus 5.00 kg u nuclei kg u λ ln T 1/ ln 9.1 yr yr min 1 R 0 λ N 0 ( min 1 )( ) counts min R e λ t 10.0 counts min R counts min and λ t ln( ) 39.6 giving t 39.6 λ yr 1 yr 45. Source: 100 mrad of -MeV γ - rays/h at a 1.00-m distance. (a) For γ - rays, dose in rem dose in rad. Thus a person would have to stand 10.0 hours to receive 1.00 rem from a 100-mrad/h source. If the γ - radiation is emitted isotropically, the dosage rate falls off as 1/r. Thus a dosage 10.0 mrad/h would be received at a distance r 10.0 m 3.16 m (a) The number of x-rays taken per year is n ( 8 x - ray d) ( 5 d wk) ( 50 wk yr) x - ray yr The average dose per photograph is 5.0 rem yr x - ray yr rem x - ray The technician receives low-level background radiation at a rate of 0.13 rem yr. The dose of 5.0 rem yr received as a result of the job is

8 Chapter 45 Solutions rem yr 38 times background levels 0.13 rem yr 000 by Harcourt, Inc. All rights reserved.

9 600 Chapter 45 Solutions 45.4 (a) I I 0 e µ x, so x 1 µ ln I 0 I With µ 1.59 cm 1 1, the thickness when I I 0 is x ln 1 ( ) cm 1.59 cm When I 0 I , x ln 1 ( ) 5.79 cm 1.59 cm rad 10 J / kg Q mc T Pt mc T t mc T P ( 50.0 C ) m 4186 J / kg C ( 10) ( 10 J / kg s)(m) s 4 days! Note that power is the product of dose rate and mass Q absorbed energy (1000 rad) 10 J / kg 10.0 J / kg m unit mass 1 rad The rise in body temperature is calculated from Q mc T where c 4186 J/kg C for water and the human body T Q mc (10.0 J / kg) J / kg C C (Negligible) 45.7 If half of the MeV gamma rays are absorbed by the patient, the total energy absorbed is MeV E g 98.9 g mol nuclei 1 mol MeV E ( MeV) ( J MeV) 0.68 J Thus, the dose received is Dose 0.68 J 1 rad 60.0 kg 10 Jkg 1.14 rad

10 Chapter 45 Solutions The nuclei initially absorbed are N g nuclei mol 89.9 g mol The number of decays in time t is At the end of 1 year, N N 0 N N 0 ( 1 e λ t ) N 0 1 e ln t 1.00 yr T yr t T 1 and N N 0 N ( )( 1 e ) The energy deposited is E ( )( 1.10 MeV) ( J MeV) J Thus, the dose received is.077 J Dose kg Jkg rad 45.9 (a) 1 E C( V) E β MeV N Q e C( V) e ( V) F MeV J / MeV ( V) F C electrons (a) amplification energy discharged E 1 C( V) E C( V) E charge released N charge of electron C V e (a) E I 10.0 ev is the energy required to liberate an electron from a dynode. Let n i be the number of electrons incident upon a dynode, each having gained energy e( V) as it was accelerated to this dynode. The number of electrons that will be freed from this dynode is N i n i e V At the first dynode, n i 1 and N 1 (1)e 100 V 10.0 ev 101 electrons E I : 000 by Harcourt, Inc. All rights reserved.

11 60 Chapter 45 Solutions For the second dynode, n i N , so N (101 )e( 100 V) ev At the third dynode, n i N 10 and N 3 (10 )e( 100 V) ev Observing the developing pattern, we see that the number of electrons incident on the seventh and last dynode is n 7 N (c) The number of electrons incident on the last dynode is n The total energy these electrons deliver to that dynode is given by E n i e( V) 10 6 e( 700 V 600 V) 10 8 ev *45.3 (a) The average time between slams is 60 min min. Sometimes, the actual interval is nearly zero. Perhaps about equally as often, it is 1.6 min. Perhaps about half as often, it is min. Somewhere around min 8.0 min, the chances of randomness producing so long a wait get slim, so such a long wait might likely be due to mischief. The midpoints of the time intervals are separated by 5.00 minutes. We use R R 0 e λ t. Subtracting the background counts, or ln [ ] ( T ) e ln. min ln ( 0.88 ) 3.47 min T 1 which yields T min. (c) As in the random events in part (a), we imagine a ± 5 count counting uncertainty. The smallest likely value for the half-life is then given by ln min T 1, or T 1 The largest credible value is found from min 1.1 min ln min T 1, yielding T Thus, T 1 ± max 38.8 min min ( 30 ± 9) min 30 min ± 30%

12 Chapter 45 Solutions The initial specific activity of 59 Fe in the steel, (R / m) 0 After 1000 h, 0.0 µci 0.00 kg 100 µci kg Bq 1 µci Bq / kg R m (R / m) 0 e λ t ( ( Bq / kg)e 4 h 1 )(1000 h) Bq / kg The activity of the oil, R oil 800 Bq / liter (6.50 liters) 86.7 Bq 60.0 Therefore, m in oil R oil (R / m) 86.7 Bq Bq / kg kg So that wear rate is h kg kg/h *45.34 The half-life of 14 O is 70.6 s, so the decay constant is λ ln ln s 1 T s The 14 O nuclei remaining after five min is N N 0 e λ t ( ( s 1) ( 300 s) )e The number of these in one cubic centimeter of blood is 1.00 cm cm 3 N N total vol. of blood cm and their activity is R λ N ( s 1 )( ) Bq ~10 3 Bq *45.35 (a) The number of photons is 10 4 MeV 1.04 MeV Since only 50% of the photons are detected, the number of 65 Cu nuclei decaying is twice this value, or In two halflives, three-fourths of the original nuclei decay, so 4 3 N and N This is 1% of the 65 Cu, so the number of 65 Cu is ~10 6. Natural copper is 69.17% 63 Cu and 30.83% 65 Cu. Thus, if the sample contains N Cu copper atoms, the number of atoms of each isotope is N N Cu and N N Cu. Therefore, N N or N N The total mass of copper present is then m Cu ( 6.93) ( ) m Cu ( 6.93 u)n 63 + ( u)n 65 : [ ( )] u gu g ~10 15 g (a) Starting with N 0 radioactive atoms at t 0, the rate of increase is (production decay) 000 by Harcourt, Inc. All rights reserved.

13 604 Chapter 45 Solutions dn dt R λ N so dn (R λ N)dt The variables are separable. N N 0 dn R λ N t t 0 dt 1 λ ln R λ N t so ln R λ N λt R R R λ N R e λ t and 1 λ R N e λ t Therefore, N R ( λ 1 e λ t ) The maximum number of radioactive nuclei would be R / λ (a) At K, each carbon nucleus has thermal energy of 3 k B T (1.5)( ev / K)( K) ev [ ] c The energy released is E m( C 1 ) m( Ne 0 ) m He 4 In the second reaction, E m( C 1 ) m Mg 4 E ( )(931.5) MeV 4.6 MeV [ ] MeV / u E ( )(931.5)MeV 13.9 MeV (c) The energy released is the energy of reaction of the # of carbon nuclei in a.00-kg sample, which corresponds to atoms / mol E g 1.0 g / mol ( ) ( 4.6) E kwh kwh 4.6 MeV / fusion event nuclei / fusion event 1 kwh MeV

14 Chapter 45 Solutions (a) Suppose each 35 U fission releases 08 MeV of energy. Then, the number of nuclei that must have undergone fission is N total release energy per nuclei J ( 08 MeV) J MeV nuclei nuclei mass g mol nuclei mol ( 35 ) kg For a typical 35 U, Q 08 MeV; and the initial mass is 35 u. Thus, the fractional energy loss is Q mc 08 MeV 35 u ( MeV u) % For the D-T fusion reaction, Q 17.6 MeV. 014 u u u The initial mass is m + The fractional loss in this reaction is Q mc 17.6 MeV 5.03 u ( MeV u) % % % 3.95 or 35 the fractional loss in D T fusion is about 4 times that in U fission To conserve momentum, the two fragments must move in opposite directions with speeds v 1 and v such that m 1 v 1 m v or v m 1 v 1 The kinetic energies after the break-up are then K 1 1 m 1 v 1 and K 1 m v 1 m m m 1 m v 1 m 1 K 1 m The fraction of the total kinetic energy carried off by m 1 is K 1 K 1 + K K 1 K 1 + ( m 1 m )K 1 m m 1 + m and the fraction carried off by m is 1 m m 1 + m m 1 m 1 + m 000 by Harcourt, Inc. All rights reserved.

15 606 Chapter 45 Solutions The decay constant is λ ln T 1 The tritium in the plasma decays at a rate of ln ( 1.3 yr) syr R λ N s cm cm 3 1 m m3 R Bq Ci Bq Bq 48 Ci s 1 The fission inventory is Ci 8 ~ 10 times greater than this amount. 48 Ci Goal Solution The half-life of tritium is 1.3 yr. If the TFTR fusion reactor contained 50.0 m 3 of tritium at a density equal to ions / cm 3, how many curies of tritium were in the plasma? Compare this value with a fission inventory (the estimated supply of fissionable material) of Ci. G : O : A : It is difficult to estimate the activity of the tritium in the fusion reactor without actually calculating it; however, we might expect it to be a small fraction of the fission (not fusion) inventory. The decay rate (activity) can be found by multiplying the decay constant λ by the number of 1 3 H particles. The decay constant can be found from the half-life of tritium, and the number of particles from the density and volume of the plasma. The number of Hydrogen-3 nuclei is particles N 50.0 m 3 m 3 λ ln yr T 1/ 1.3 yr s s -1 The activity is then 100 cm particles m R λn ( s -1 )( nuclei) Bq ( Bq) 1 Ci Bq 48 Ci L : Even though 48 Ci is a large amount of radioactivity, it is smaller than Ci by about a hundred million. Therefore, loss of containment is a smaller hazard for a fusion power reactor than for a fission reactor.

16 Chapter 45 Solutions Momentum conservation: 0 m Li v Li + m α v α, or, m Li v Li m α v α Thus, K Li 1 m Li v Li 1 m Li v Li m Li ( m α v α ) m Li m α m Li v α u K Li u K Li kg ms 1.14 u ms ms J 1.0 MeV The complete fissioning of 1.00 gram of U 35 releases (1.00 g) Q atoms 35 grams / mol mol 00 MeV J fission MeV J If all this energy could be utilized to convert m kilograms of 0.0 C water to 400 C steam (see Chapter 0 of text for values), then Q mc w T + ml v + mc s T [ 80.0 C ( 300 C) ] Q m 4186 J / kg C J / kg J / kg C Therefore m J J / kg kg When mass m of 35 U undergoes complete fission, releasing 00 MeV per fission event, the total energy released is: m Q 35 g mol N A 00 MeV where N A is Avogadro s number. If all this energy could be utilized to convert a mass m w of liquid water at T c into steam at T h, then, [ ] Q m w c w ( 100 C T c )+ L v + c s T h 100 C where c w is the specific heat of liquid water, L v is the latent heat of vaporization, and c s is the specific heat of steam. Solving for the mass of water converted gives m w Q [ c w ( 100 C T c )+ L v + c s ( T h 100 C) ] mn A ( 00 MeV) [ ] ( 35 g mol) c w ( 100 C T c )+ L v + c s T h 100 C 000 by Harcourt, Inc. All rights reserved.

17 608 Chapter 45 Solutions (a) The number of molecules in 1.00 liter of water (mass 1000 g) is N g 18.0 g mol ( molecules mol) molecules The number of deuterium nuclei contained in these molecules is N molecules 1 deuteron 3300 molecules deuterons Since deuterons are consumed per fusion event, the number of events possible is N reactions, and the energy released is E fusion ( reactions) ( 3.7 MeV reaction) MeV E fusion ( MeV) ( J MeV) J In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is E fusion J E gasoline times larger. J The number of nuclei in kg of 10 Po is 155 g N g mol ( nuclei g) nuclei The half-life of 10 Po is days, so the decay constant is given by λ ln T 1 ln ( d) sd s 1 The initial activity is R 0 λ N 0 ( s 1 )( nuclei) Bq The energy released in each Po 8Pb + He reaction is Q M 10 84Po M 06 8 Pb M 4 He c : Q [ ]u MeV 5.41 MeV u Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is P ( )R 0 Q decays s MeV J 5.41 decay MeV 3 W

18 Chapter 45 Solutions (a) The thermal power transferred to the water is P w 0.970( waste heat) r w is the mass of heated per hour: r w P w c T P w 0.970( )MW Js 3600 s h ( 3.50 C) Js 4186 J kg C kg h The volume used per hour is kg h kg m m h The 35 U fuel is consumed at a rate r f Js 1 kg Jg 1000 g 3600 s kg/h 1 h *45.48 (a) V 4π r ( r) 4π ( m) ( 0.05 m) m 3 ~10 8 m 3 The force on the next layer is determined by atmospheric pressure. W P( V) N ( m 3 ) J ~10 J (c) J 10 1 ( yield ), so yield J ~10 J m (d) J J ton TNT 4 4 ton TNT ~ 10 ton TNT or ~ 10 kilotons *45.49 (a) V l 3 m ρ, so l m ρ kg kg m m Add 9 electrons to both sides of the given nuclear reaction. Then it becomes 38 9U atom 8 4 He atom + 8 Pb atom + Q net. Q net M 38 9U 8 M 4 He M 8 Q net 51.7 MeV 06 Pb c [ ( ) ]u MeV u (c) If there is a single step of decay, the number of decays per time is the decay rate R and the energy released in each decay is Q. Then the energy released per time is P QR. If there is a series of decays in steady state, the equation is still true, with Q representing the net decay energy. 000 by Harcourt, Inc. All rights reserved.

19 610 Chapter 45 Solutions (d) The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: N g 38 g mol ( nuclei mol) nuclei λ ln ln T yr yr R λ N yr ( nuclei) decays yr, so P QR 51.7 MeV yr J MeV. Jyr (e) dose in rem dose in rad x RBE rem 500. dose in rad 110. yr yr, giving dose in rad yr 455. rad yr 10 The allowed whole-body dose is then kg rad Jkg yr 1 rad 3.18 J/yr E T E( thermal) 3 k BT ev E T ( 1 ) n E where n number of collisions, and ( 1 ) n ( ) Therefore, n collisions From conservation of energy: K α + K n Q or 1 m α v α + 1 m n v n 17.6 MeV Conservation of momentum: m α v α m n v n or v α m n v n. The energy equation becomes: 1 m m n α v n + 1 m α m n v n m n + m α m α ( 1 m n v n ) 17.6 MeV m Thus, K n α ( 17.6 MeV) m n + m α MeV m α

20 Chapter 45 Solutions 611 Goal Solution Assuming that a deuteron and a triton are at rest when they fuse according to H + 3 H 4 He + n MeV, determine the kinetic energy acquired by the neutron. G : The products of this nuclear reaction are an alpha particle and a neutron, with total kinetic energy of 17.6 MeV. In order to conserve momentum, the lighter neutron will have a larger velocity than the more massive alpha particle (which consists of two protons and two neutrons). Since the kinetic energy of the particles is proportional to the square of their velocities but only linearly proportional to their mass, the neutron should have the larger kinetic energy, somewhere between 8.8 and 17.6 MeV. O : Conservation of linear momentum and energy can be applied to find the kinetic energy of the neutron. We first suppose the particles are moving nonrelativistically. A : The momentum of the alpha particle and that of the neutron must add to zero, so their velocities must be in opposite directions with magnitudes related by m n v n + m α v α 0 or ( u)v n ( u)v α At the same time, their kinetic energies must add to 17.6 MeV E 1 m n v n + 1 m α v α 1 ( u )v n + 1 ( u )v α 17.6 MeV Substitute v α 0.50v n : E ( u)v n + ( u)v 1 u n 17.6 MeV MeV / c v n c c m/s Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic energy of the neutron from the classical equation, K 1 mv u ( 0.173c MeV / c ) u 14.1 MeV L : The kinetic energy of the neutron is within the range we predicted. For a more accurate calculation of the kinetic energy, we should use relativistic expressions. Conservation of momentum gives γ n m n v n + γ α m α v α yielding v α v n 1 v n / c v α 1 v α / c c v n c v n Then ( γ n 1)m n c + ( γ α 1)m α c 17.6 MeV and v n 0.171c, implying that ( γ n 1)m n c 14.0 MeV 000 by Harcourt, Inc. All rights reserved.

21 61 Chapter 45 Solutions 45.5 From Table A.3, the half-life of 3 P is 14.6 d. Thus, the decay constant is λ ln ln T d d s 1. N 0 R 0 λ decay s s nuclei At t 10.0 days, the number remaining is N N 0 e λ t ( ( d 1) ( 10.0 d) nuclei)e nuclei so the number of decays has been N 0 N and the energy released is E ( ) 700 kev J kev J If this energy is absorbed by 100 g of tissue, the absorbed dose is J rad Dose kg 0 10 Jkg 400 rad (a) The number of Pu nuclei in 1.00 kg nuclei/mol g/mol (1000 g) The total energy ( nuclei)(00 MeV) MeV E ( MeV)( kwh/mev) kwh or million kwh E mc ( u u u u) (931.5 MeV/u) E 17.6 MeV for each D-T fusion (c) E n (Total number of D nuclei)(17.6)( ) E n ( )(1000/.014)(17.6)( ) kwh (d) E n the number of C atoms in 1.00 kg 4.0 ev E n ( /1.0)( MeV)( ) 9.36 kwh (e) Coal is cheap at this moment in human history. We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels.

22 Chapter 45 Solutions 613 *45.54 Add two electrons to both sides of the given reaction. Then H atom 4 He atom + Q where or Q ( m)c [ 4( ) ]u ( MeV u) 6.7 MeV Q ( 6.7 MeV) ( J MeV) J The proton fusion rate is then rate power output energy per proton Js ( J) 4 protons protons s *45.55 (a) Q I [ M A + M B M C M E ]c, and Q II [ M C + M D M F M G ]c Q net Q I + Q II [ M A + M B M C M E + M C + M D M F M G ]c Q net Q I + Q II [ M A + M B + M D M E M F M G ]c Thus, reactions may be added. Any product like C used in a subsequent reaction does not contribute to the energy balance. Adding all five reactions gives 1 1 H H e H H e 4 He + ν + Q net or Adding two electrons to each side H e 4 He + ν + Q net H atom 4 He atom + Q net [ 1 ] c Thus, Q net 4 M 1 H M 4 He [ ]u ( MeV u) 6.7 MeV (a) The mass of the pellet is m ρv 0.00 g cm 3 4π cm g The pellet consists of equal numbers of H and 3 H atoms, so the average atomic weight is.50 and the total number of atoms is N g.50 g mol ( atoms mol) atoms When the pellet is vaporized, the plasma will consist of N particles ( N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 00 kj or.00 kj. The temperature of the plasma is found from E N T E 3Nk B J ( JK) as 3 k B T K Each fusion event uses nuclei, so N events will occur. The energy released will be E N Q MeV J MeV J 10 kj 000 by Harcourt, Inc. All rights reserved.

23 614 Chapter 45 Solutions *45.57 (a) The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the Coulomb-repulsion barrier to 1 H+ He He+ e + ν, estimated as k e ()e e barrier to Bethe s fifth and eight reactions is like k e ()7e e r, larger by 7 temperature should be like ( K ) K. For 1 C + 1 H 13 N + Q, Q 1 ( ) ( MeV) 1.94 MeV 4 r. The Coulomb times, so the For the second step, add seven electrons to both sides to have: 13 N atom 13 C atom + e + e + + Q. Q [ ( ) ]( MeV) 1.0 MeV Q 3 Q 7 ( ) ( MeV) 1.0 MeV Q 4 [ ] ( MeV) 7.55 MeV Q 5 [ ] ( MeV) 7.30 MeV Q 6 [ ( ) ]( MeV) 1.73 MeV Q 8 [ ] ( MeV) 4.97 MeV The sum is 6.7 MeV, the same as for the proton-proton cycle. (c) Not all of the energy released heats the star. When a neutrino is created, it will likely fly directly out of the star without interacting with any other particle (a) (c) I e µ x x I 0 I 1 I 0 e µ 1 x e µ µ 1 I 50 e ( )( 0.100) e I 100 I 50 e ( )( 1.00) e I 100 Thus, a 1.00-cm aluminum plate has essentially removed the long-wavelength x-rays from the beam.