Chapter 30 Questions 8. Quoting from section 30-3, K radioactivity was found in every case to be unaffected

Size: px

Start display at page:

Download "Chapter 30 Questions 8. Quoting from section 30-3, K radioactivity was found in every case to be unaffected"

Hugo Welch
5 years ago
Views:

1 Physics 111 Fall 007 Homework Solutions Week #10 Giancoli Chapter 30 Chapter 30 Questions 8. Quoting from section 30-3, K radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong chemical reagents. Chemical reactions are a result of electron interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to electron interactions. Another piece of evidence is the fact that the -particle is much heavier than an electron and has a different charge than the electron, so it can t be an electron. Therefore it must be from the nucleus. Finally, the energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron orbital transitions. All of these observations support radioactivity being a nuclear process. 11. Gamma rays are neutrally charged, they are made up of high energy photons (travel at c, and they have no mass. Alpha rays are made up of helium nuclei, they are the most massive of these three particles, and they have a charge of +e. Beta rays are made up of electrons or positrons, they can be either positively or negatively charged, and the particles are accompanied by either a neutrino or an anti-neutrino upon decay. 1. (a 4 Na Mg + e + v Magnesium-4 is formed. (b 4 Na Ne + e + + v Neon- is formed.

2 (c 10 Po 4 84 He Pb Lead-06 is formed. 13. (a 3 P S+ e + v Sulfur-3 is formed. (b 35 S Cl + e + v Chlorine-35 is formed. (c 11 Bi 4 83 He T1 Thallium-07 is formed. 14. (a 45 Ca Sc + e + v Scandium-45 is the missing nucleus. (b (c (d 58 Cu Cu + γ 46 Cr V + e + + v particles. 34 Pu U +α Copper-58 is the missing nucleus. The positron and the neutrino are the missing Uranium-30 is the missing nucleus. (e 39 Np Pu + e + v The electron and the anti-neutrino are the missing particles. 15. The two extra electrons are no longer bound tightly to the nucleus (since the nucleus lost two positively charged protons, and so those extra electrons can move away from the nucleus, moving towards a higher electric potential or some other positively charged object. They are not emitted during the decay, since they do not receive any of the kinetic energy from the decay. If they are close to each other, they would repel each other somewhat. Problems 1. We convert the units: m = ( 139 MeV c = u MeV uc

3 5. (a The mass of a nucleus with mass number A is A u and its radius is r = ( ma 1 3. Thus the density is ρ = m V = A kg u = A 4 3 πr3 ( kg u 3 A 4 π m = kg m 3, independent of A. (b We find the radius from M = pv ; 4 3 πr 3, which gives R = 180 m kg = kg m 3 (c For equal densities, we have ρ = M Earth 4 3 πr Earth = m U 3 4 πr ; 3 3 U ( kg ( ( m = 38 u1.66 ( 10 7 kg u, which gives r 3 3 U = m. r U 13. We find the binding energy of the last neutron from the masses: Binding energy= m( 39 K+ m( 1 n m( 40 K c = ( u+ ( u ( u c MeV c = MeV.. For each decay, we find the difference of the initial and the final masses: m = m( 36 U m( 35 U m( 1 n (a = ( u ( u ( u= u. (b Because an increase in mass is required, the decay is not possible. m = m( 16 O m( 15 O m( 1 n = ( u ( u ( u= u.

4 (c Because an increase in mass is required, the decay is not possible. m = m( 3 Na m( Na m( 1 n = ( u ( u ( u= u. Because an increase in mass is required, the decay is not possible. 4. The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the sodium, the maximum kinetic energy of the electron is KE = m( 3 Ne m( 3 Na c = (.9945 u ( u ( c MeV uc = 4.4 MeV. When the neutrino has all of the kinetic energy, the minimum kinetic energy of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be from the mass difference, so the energy range of the neutrino will be 0 E v 4.4 MeV. 33. If we ignore the recoil of the lead, the kinetic energy of the α particle is KE = m( 10 Po m( 16 Pb m( 4 He c = ( u ( u ( u ( c MeV uc = 5.41 MeV. 35. For the positron-emission process A A χ Z χ + e + + v, Z +1 we need to add Z + 1 electrons to the nuclear mass of χ to be able to use the atomic mass. On the right-hand side we use Z electrons to be able to use the atomic mass of χ. Thus we have 1 electron mass and the β-particle mass, which means that we must include electron masses on the right-hand side. The Q value will be Q = M P ( M D + m e c = ( M p M D m e c. 36. We find the decay constant from

5 λn = λn 0 e λt ; 30 decays min = ( 180 decays mine Thus the half-life is, which gives λ = h. λ 4.6 h = λ = =.3 h h 39. We find the fraction remaining from N = N 0 e λt ; N N 0 = e λt = e ( ( yr1 ( mo yr 9 mo = The decay constant is λ = The initial number of nuclei is = ( 8.0 days4 ( h day3600 s h = s g N 0 = ( atoms mol= nuclei. 131 g mol (a When t = 0, we get λn = λn 0 e λt = ( s ( e 0 = decays s. (b When t =1.0 h, the exponent is so we get λt = ( s ( h ( 3600s h= , λn = λn 0 e λt = ( s ( e = decays s. (c When t = 6 months, the exponent is λt = ( s 1 6 ( mo ( 30.5days mo ( 4h day ( 3600 s h= 15.81,

6 so we get λn = λn 0 e λt = ( s ( e = decays s. 49. We find the number of half-lives from N t = ( 1 n ; N t = ( 1 n, or n log = log 10, which gives n = 3.3. Thus the half-life is = t 8.6 min = =.6 min. n We assume that the elapsed time is much smaller than the half-life, so we can use a constant decay rate. Because 87 Sr is stable, and there was none present when the rocks were formed, every atom of 87 Rb that decayed is now an atom of 87 Sr. Thus we have N Sr = N Rb = λn Rb t, or N Sr = t; N Rb T = yr t, which gives t = yr. This is % of the half-life, so our original assumption is valid. 56. The decay constant is λ = = ( yr = /yr. Because the fraction of atoms that are C is so small, we use the atomic weight of C to find the number of carbon atoms in 90 g:

7 90 g N = ( 1 g mol atoms mol= atoms, so the number of 14 C nuclei in a sample from a living tree is N 14 = ( ( 10 5 = nuclei. Because the carbon is being replenished in living trees, we assume that this number produced the activity when the club was made. We determine its age from λn = λn 14 e λt ; 4 ( yr t, 8.0 decays s = yr ( nucleie s yr which gives t = yr. 57. The number of radioactive nuclei decreases exponentially: N = N 0 e λt. Every radioactive nucleus that decays becomes a daughter nucleus, so we have N D = N 0 N = N 0 ( 1 e λt.

NJCTL.org 2015 AP Physics 2 Nuclear Physics

NJCTL.org 2015 AP Physics 2 Nuclear Physics AP Physics 2 Questions 1. What particles make up the nucleus? What is the general term for them? What are those particles composed of? 2. What is the definition of the atomic number? What is its symbol?