Radioactivity Solutions - Lecture 28B (PHY315)

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1 Radioactivity s - Lecture 8B (PHY35) Problem solutions.strategy In beta-minus decay, the atomic number Z increases by while the mass number A remains constant. Use Eq. (9-). 4 For the parent 9 K Z 9, so the daughter nuclide will have Z 9 +, which is the element Ca. he symbol for the daughter is 4 Ca.. Strategy In alpha decay, the atomic number Z and the mass number A are decreased by and 4, respectively. Use (Eq. 9-). In this case, the parent nuclide has Z 9 and A 3, so the daughter nuclide will have Z 88 and A 8, which is the element radium. Write out the reaction. A Z A Z, so 9h 88Ra. he daughter nuclide is P D 8 88 Ra. 3. Strategy In electron-capture decay, the atomic number Z is decreased by while the mass number A stays the same. In this case, the parent nuclide has Z and A, so the daughter nuclide will have Z and A, which is the element neon. Write out the reaction. A Z A e Z, so Na e Ne. he daughter nuclide is P D Ne. 4. Strategy In beta-plus decay a positron is emitted, the atomic number Z is decreased by while the mass number A stays the same. Use Eq. (9-). In this case, the parent nuclide has Z and A, so the daughter nuclide will have Z and A, which is the element neon. Write out the reaction. A Z A Z e, so Na Ne e. he daughter nuclide is P D Ne. 5. Strategy he kinetic energy of the decay products equals the energy associated with the change in mass. Find the change in mass during the decay using atomic masses. 4 m (mass of 8Rn mass of He) (mass of 88Ra) (. 5 5 u 4. 3 u).5 4 u u Find the kinetic energy. K E m c u MeV u 4.8 MeV Assuming the 8 Rn nucleus takes away an insignificant fraction of the kinetic energy, the alpha particle s kinetic energy will be 4.8 MeV.. Strategy he kinetic energy of the decay products equals the energy associated with the change in mass. Find the ratio of the kinetic energies using nonrelativistic formulas.

2 Find the change in mass during the decay using atomic masses. 4 m (mass of 8Rn mass of He) (mass of 88Ra) (. 5 5 u 4. 3 u).5 4 u u Find the kinetic energy. K E m c u MeV u 4.8 MeV he momentum before the decay was, so conservation of momentum requires that the momenta of the decay products must be equal in magnitude and opposite in direction (for a total momentum of ). Compute the ratio of the kinetic energies of the alpha particle and the daughter nuclide. K p ( m ) md. 5 5 u K d p ( md ) m 4. 3 u he sum of the two kinetic energies must equal 4.8 MeV. K 4.8 MeV K Kd K 4.8 MeV, so K MeV he rest energy of an alpha particle is E m c 4. 3 u MeV u 38.4 MeV. Since K E, the alpha particle is indeed nonrelativistic.. Strategy Large nuclides with relatively low neutron-to-proton ratios are like to decay via alpha decay. Smaller nuclides with too many neutrons are likely to decay via beta-minus decay and those with too few neutrons via beta-plus decay. his nuclide is relatively small. he neutron to proton ratio is (34) 4.. his ratio is low, so we would expect that the nuclide has too many neutrons. Also, stable isotopes of silicon have fewer neutrons than 3 4 Si, so the expected decay should convert a neutron into a proton. decay has the desired effect. 8. Strategy In a spontaneous decay, the mass of the decay products must be less than the mass of the parent. Alpha decay of 9 8 O would have a daughter nuclide of 5 C O C He he mass of atomic 9 8 O is u. he combined mass of 5 C and 4 He is u u u. Since u > u, the spontaneous alpha decay of 9 8 O is not possible. 9. Strategy In Problem, the daughter nuclide in this decay was found to be 4 Ca. he maximum kinetic energy of the beta particle is equal to the disintegration energy. he reaction is K Ca e he atomic masses of K and Ca are u and u, respectively. o get the masses of the nuclei, we subtract Zm e from each. he mass of the electron is. 548 u, and the neutrino s mass is negligible. he mass difference is m [( MCa me) me] ( MK 9 me) MCa MK u u.4 5 u.

3 he disintegration energy is E m c. 4 5 u MeV u.3 MeV. he maximum kinetic energy of the particle is.3 MeV. 3. Strategy In beta-minus decay an electron is emitted; the atomic number Z increases by while the mass number A stays the same. Use Eq. (9-) and conservation of energy to find the energy of the antineutrino. In this case, the parent nuclide has Z 38 and A 9, so the daughter nuclide will have A 9 and Z , which is the element yttrium. Write out the reaction. A A 9 9 ZP ZD e, so 38Sr 39Y e. he atomic masses of strontium-9 and yttrium-9 are u and u, respectively. o get the masses of the nuclei, we subtract Zm e (the mass of each atom s electrons) from the atomic masses. he neutrino s mass is negligible. Calculate the mass difference. m [( MY 39 me ) me ] ( MSr 38 me ) MY MSr u u. 58 u he disintegration energy is E m c. 58 u MeV u 54. kev. Of this energy released by the decay, 435 kev is the kinetic energy of the beta particle and 54. kev 435 kev kev is the energy of the antineutrino. 3. Strategy In beta-plus decay a positron is emitted; the atomic number Z is decreased by while the mass number A stays the same. Use Eq. (9-) and conservation of energy to estimate the maximum possible kinetic energy of the emitted positron. In this case, the parent nuclide has Z and A, so the daughter nuclide will have Z and A, which is the element neon. Write out the reaction. A A ZP ZD e, so Na Ne e. he atomic masses of sodium- and neon- are u and u, respectively. o get the masses of the nuclei, we subtract Zm e (the mass of each atom s electrons) from the atomic masses. he neutrino s mass is negligible. Calculate the mass difference. m [( MNe me ) me ] ( MNa me ) M Ne M Na me u u ( u).954 u he disintegration energy is E m c u MeV u.83 MeV. Assuming that the neon nucleus has nearly zero kinetic energy and the total energy of the neutrino is zero, the maximum possible kinetic energy of the emitted positron is about.8 MeV. 3. Strategy In electron-capture decay, the atomic number Z is decreased by while the mass number A stays the same. Use conservation of energy and momentum. Use the classical expression for the kinetic energy of the carbon atom and the extremely relativistic expression for the kinetic energy of the neutrino. In this case, the parent nuclide has Z and A, so the daughter nuclide will have Z and A, which is the element carbon. Write out the reaction. A e A ZP ZD, so N e C. he atomic masses of nitrogen- and carbon- are.8 3 u and. u, respectively. o get the masses of the nuclei, we subtract Zm e (the mass of each atom s electrons) from the atomic masses. he neutrino s mass is negligible. Calculate the mass difference.

4 m ( M m ) [( M m ) m ] M M. u.8 3 u.8 3 u C e N e e C N he disintegration energy is E m c.8 3 u MeV u.338 MeV. his energy is shared between the neutrino and the carbon nucleus. Conservation of energy gives pc E KC E E. M Conservation of momentum gives p C C p E c. Combining these two equations gives E E E, or. MCc MCc Using the quadratic formula to solve for E results in E E E E.35 MeV. 33. Strategy he activity is reduced by a factor of two for each half-life. Use Eqs. (9-8), (9-), and (9-) to find the initial number of nuclei and the probability of decay per second. (a) Find the number of half-lives.. s. s 3. half-lives. s half-life he activity after 3. half-lives will be (b) Find the initial number of nuclei. R /. s N R R 8,. s.38 ln ln 3. R R 8,. s, s (c) he probability per second is ln ln 3.4 s. /. s 34. Strategy Use Eq. (9-) to find the time constant. he number of nuclei is related to the mass by N mna M, where NA is Avogadro s number and M is the molar mass. he activity is equal to the number of nuclei divided by the time constant. Convert the half-life of radium- to seconds. 3.5 s yr 5.49 s yr he time constant is / 5.49 s.85 s. ln ln Find the number of nuclei in. g of radium. mna. g 3 N. nuclei mol.43 nuclei M.54 g mol N.43 nuclei Ci he activity is R.99 Ci..85 s 3. Bq 35. Strategy Use Eq. (9-) to find the time constant. he number of nuclei is related to the mass by N mna M, where NA is Avogadro s number and M is the molar mass. he activity is equal to the number of nuclei divided by the time constant.

5 Convert the half-life of uranium-38 to seconds yr 3.5 s yr.4 s he time constant is /.4 s.34 s. ln ln Find the number of nuclei in. kg of 38 U. 3 mna. g 3 4 N. nuclei mol.5 nuclei M 38.5 g mol he activity is 4 N.5 nuclei R. Bq..34 s t 3. Strategy he activity as a function of time is given by R R e. Use Eq. (9-) to find the time constant. Find the number of days for the activity to decrease to.5 Bq. t R t R R 8. d.5 Bq e, so ln or t ln ln 4 d. R 8 R R ln.4 Bq t 3. Strategy he activity as a function of time is given by R R e. Use Eq. (9-) to find the time constant. Assume that the original activity was.5 Bq per g of carbon. Find the age of the bones. t R t R R 53 yr.4 e, so ln or t ln ln yr. R R R ln Strategy Use Eqs. (9-) and (9-) to find the decay constant. he total number of nuclei in. g of carbon times the relative abundance gives the number of carbon-4 nuclei. Use the results of parts (a) and (b) and Eq. (9-8) to find the activity per gram in a living sample. (a) Calculate the decay constant. ln ln 3.83 s / 53 yr 3.5 s yr (b) Calculate the number of 4 C atoms. mass. g 3 N NA (relative abundance). atoms mol.3 mass per mol. g mol.5 atoms (c) Calculate the activity per gram. R N.5 atoms 3.83 s.5 Bq/g. g. g. g

6 39. Strategy he ratio of C-4 to C- in the bone is /4 as much as in a living sample. he ratio is reduced by a factor of / for each half-life. Since 4, we conclude that the age of the bone is half-lives, or 53 yr,5 yr. 4. Strategy he half-life of 4 83 Bi is 9.9 min. he activity is given by R R / ( t ). Find the activity. /9.9 t R R ( / ).58 Ci. Ci t tln 4. Strategy he activity is given by R R / e R e. Solve for the half-life. tln R e / R tln R ln / R t ln min ln / 3 ln R. Bq R ln 4.4 Bq.4 min 4. Strategy Look up the half-lives in Appendix B. Use Eq. (9-3). he numbers of oxygen-5 nuclei and oxygen-9 nuclei at time t are given by N 5 5 N t and 9 N 9, t N 5 t N 9 t 5 t 9 N respectively. here are twice as many oxygen-5 nuclei as oxygen-9 nuclei at time t. Find t. t t t (.4 s)(.88 s) t.4 s.88 s 34.4 s 5 9 Compute the percent of oxygen-9 nuclei that have decayed. N N9 t % ( ) % ( ) % 58.8% N 43. Strategy ake the logarithm of each side of the given equation and simplify to prove the assertion. First, we see that Solve for /. t / t / t/ t ( ) is always true. Next, we start with / t e.

7 t ln( / t ) ln( e ) t t ln / ln / ln / Since each step in the demonstration is reversible, we conclude that / ln. t t/ / t e if and only if 44. Strategy Use the rules of probability and the definition of half-life. (a) (expected # of decays) (probability of throwing a one ) (number of dice thrown) N N (b) he probability that a single die is undecayed after a single toss is 5. If N dice are tossed once, the expected number of undecayed dice is 5 N. After two tosses, the expected number of undecayed dice is N N. After three tosses the number is 3 5 N. (c) Following the pattern in part (b), the answer is 4 5 N. (d) After one half-life, /, N the number of undecayed dice would be. he number of undecayed dice is t 5 N given by Nundecayed N, where t is the number of tosses. Set Nundecayed and solve for / t. / N 5 N / 5 5 / ln ln ln / 3.8 tosses ln 5

General Physics (PHY 2140)

General Physics (PHY 2140) Lecture 37 Modern Physics Nuclear Physics Radioactivity Nuclear reactions http://www.physics.wayne.edu/~apetrov/phy2140/ Chapter 29 1 Lightning Review Last lecture: 1. Nuclear