Physics Solution Manual for 1100 and 2101

After calculating the mass difference in atomic mass

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Unformatted text preview: y 4. Because the atomic mass number of the daughter nucleus, lead 207 Pb , is 207, we have that 82 A − 4 = 207 or A = 207 + 4 = 211 Similarly, given that the atomic number of the daughter nucleus is 82, the atomic number Z of the parent nucleus must be Z = 82 + 2 = 84. This atomic number identifies the parent nucleus as polonium (Po). Therefore, the decay process is 211 207 4 84 Po → 82 Pb + 2 He . b. In a β − decay, the atomic mass number A is unaffected, but the atomic number Z increases by 1. Therefore, the parent nucleus in this decay process has an atomic number that is 1 less than that of the daughter nucleus: Z = 82 − 1 = 81. Thallium (Tl) is the element with the atomic number 81, so we have that 207 207 0 81Tl → 82 Pb + −1e . 25. REASONING We can determine the identity of X in each of the decay processes by noting that for each process, the sum of A and Z for the decay products must equal the values of A and Z for the parent nuclei. SOLUTION a. 211 82 Pb → 211 Bi + X 83 Using the reasoning discussed above, X must have A = 0 and Z = –1 . Therefore X must be an electron, 0 –1 e; X represents a β – particle (electron) . 11 6 b. C → 11 B + X 5 Similar reasoning suggests that X must have A = 0 and Z = +1 . Therefore X must be a positron; 0 +1 e; X represents a β + particle (positron) . 1600 NUCLEAR PHYSICS AND RADIOACTIVITY 231 90 c. Th * → 231 Th + X 90 Similar reasoning suggests that X must have A = 0 and Z = 0 . Therefore X must be a gamma ray; X represents a γ ray . 210 84 d. Po → 206 Pb + X 82 Using the reasoning discussed above, X must have A = 4 and Z = 2 . Therefore X must be a helium nucleus, 4 He ; X represents an α particle (helium nucleus) . 2 26. REASONING The γ-ray photon is emitted when the nucleus changes from one energy state to a lower energy state. The energy of the photon is the difference ∆E between the two nuclear energy levels, in a way similar to that discussed in Section 30.3 for the energy levels of the electron in the hydrogen atom. In that section, we saw that ∆E is related to the frequency f and Planck’s constant h. Thus, we can determine the energy from a value for the frequency. However, this value is not given. Instead, the wavelength λ is given. The frequency can be obtained from the wavelength, since the two quantities are related to the speed c of light. SOLUTION Section 30.3 discusses the fact that the photon emitted when the electron in a hydrogen atom changes from a higher to a lower energy level has an energy ∆E, which is the difference between the energy levels. A similar situation exists here when the nucleus changes from a higher to a lower energy level. The γ-ray photon that is emitted has an energy ∆E given by (30.4) ∆E = h f The frequency is related to the wavelength according to f= c (16.1) λ Substituting this expression into Equation 30.4 shows that ( c ∆ E = h f = h = 6.63 × 10−34 J ⋅ s λ 3.00 ×10 ) 1.14 ×10−11m/s = 1.74 ×10−14 J m 8 Finally, we convert this result from joules (J) to MeV: ( ) 1 eV 1 MeV ∆ E = 1.74 ×10−14 J = 0.109 MeV −19 J 1×106 eV 1.60 ×10 Chapter 31 Problems 1601 27. REASONING Since energy is released during the decay, the combined mass of the lead 206 210 82 Pb daughter nucleus and the α particle is less than the mass of the polonium 84 Po parent nucleus. The difference in mass is equivalent to the energy released. Since the recoil of the lead nucleus is being ignored and we are assuming that all the released energy goes into the kinetic energy KE of the α particle, it follows that released energy = KE = 1 mv 2 2 (Equation 6.2). Thus, the speed of the α particle is v = 2 ( KE ) . m SOLUTION The decay reaction is 210 Po 1 84 3 4 24 209.982 848 u → 206 4 He 82 Pb3 + 1224 43 14 24 4.002 603 u 205.974 440 u The difference in the masses is 209.982 848 u − 205.974 440 u − 4.002 603 u = 5.805 × 10−3 u This mass difference corresponds to an energy of (5.805 ×10−3 u ) 931.5uMeV = 5.407 MeV 1 This energy is the kinetic energy of the α particle (mass m = 6.6447 × 10–27 kg, see Example 2), so the speed of the α particle is v= 2 ( KE ) = m 1.60 × 10−19 J 2 5.407 × 106 eV 1 eV = 1.61 × 107 m/s 6.6447 × 10−27 kg ( ) 28. REASONING When a nucleus undergoes an α decay, it emits one α particle that contains two protons and two neutrons. Therefore, each α decay decreases the atomic mass number A by four. A β − decay does not affect the atomic mass number A at all, so the difference Ad − Ap between the atomic mass number Ad = 208 of the daughter nucleus atomic mass number Ap = 230 of the parent nucleus ( 208 Pb ) and the 82 ( 230 Rn ) after the emission of Nα alpha 86 particles is Ad − Ap = −4 Nα (1) 1602 NUCLEAR PHYSICS AND RADIOACTIVITY The negative sign in Equation (1) indicates a reduction in the atomic mass number in going from the parent to the daughter. In order to determine the difference Zd − Zp between the final atomic number Zd = 82 of the daughter nucleus and the atomic number Zp = 86 of the parent nucleus, we must take into account both the loss...
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