Nuclei

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Exercise 13.1Nuclear Mass, Binding Energy and Radioactive Decay

Two stable isotopes of lithium ⁶Li and ⁷Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

Solution

Atomic mass = weighted average of isotope masses: M = (7.5/100) × 6.01512 + (92.5/100) × 7.01600 = 0.075 × 6.01512 + 0.925 × 7.01600 = 0.45113 + 6.4898 = 6.94093 u ≈ 6.941 u The atomic mass of lithium is approximately 6.941 u.

Boron has two stable isotopes, ¹⁰B and ¹¹B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of ¹⁰B and ¹¹B.

Solution

Let abundance of ¹⁰B = x%, so ¹¹B = (100 – x)% (x/100) × 10.01294 + ((100–x)/100) × 11.00931 = 10.811 10.01294x + 11.00931(100–x) = 1081.1 10.01294x + 1100.931 – 11.00931x = 1081.1 –0.99637x = 1081.1 – 1100.931 = –19.831 x = 19.831/0.99637 ≈ 19.9 Abundance of ¹⁰B ≈ 19.9% Abundance of ¹¹B ≈ 100 – 19.9 = 80.1%

The three stable isotopes of neon: ²⁰Ne, ²¹Ne and ²²Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Solution

M = (90.51/100) × 19.99 + (0.27/100) × 20.99 + (9.22/100) × 21.99 = 0.9051 × 19.99 + 0.0027 × 20.99 + 0.0922 × 21.99 = 18.1030 + 0.0567 + 2.0274 = 20.1871 u ≈ 20.18 u

Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given: m(¹⁴₇N) = 14.00307 u.

Solution

¹⁴₇N has Z = 7 protons and N = 7 neutrons m_p = 1.00783 u, m_n = 1.00867 u Mass of constituents: M_constituents = 7 × 1.00783 + 7 × 1.00867 = 7.05481 + 7.06069 = 14.11550 u Mass defect: Δm = M_constituents – M_nucleus = 14.11550 – 14.00307 = 0.11243 u Binding energy: EB = Δm × 931.5 MeV/u = 0.11243 × 931.5 = 104.72 MeV Binding energy per nucleon = 104.72/14 = 7.48 MeV

Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi in units of MeV from the following data: m(⁵⁶Fe) = 55.934939 u, m(²⁰⁹Bi) = 208.980388 u.

Solution

m_p = 1.007825 u, m_n = 1.008665 u For ⁵⁶₂₆Fe (Z=26, N=30): Mass of constituents = 26 × 1.007825 + 30 × 1.008665 = 26.20345 + 30.25995 = 56.46340 u Δm = 56.46340 – 55.934939 = 0.528461 u EB = 0.528461 × 931.5 = 492.28 MeV EB/A = 492.28/56 = 8.79 MeV/nucleon For ²⁰⁹₈₃Bi (Z=83, N=126): Mass of constituents = 83 × 1.007825 + 126 × 1.008665 = 83.64948 + 127.09179 = 210.74127 u Δm = 210.74127 – 208.980388 = 1.760882 u EB = 1.760882 × 931.5 = 1640.3 MeV EB/A = 1640.3/209 = 7.85 MeV/nucleon

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to completely disintegrate all the ⁶³Cu nuclei in the coin. Given: m(⁶³Cu) = 62.92960 u.

Solution

Mass of coin = 3.0 g, M(⁶³Cu) = 63 g/mol (approximately) Number of ⁶³Cu nuclei: N = (3.0/63) × 6.022 × 10²³ = 2.868 × 10²² For ⁶³Cu (Z=29, N=34): m_p = 1.007825 u, m_n = 1.008665 u Mass of constituents = 29 × 1.007825 + 34 × 1.008665 = 29.22693 + 34.29461 = 63.52154 u Mass defect per nucleus: Δm = 63.52154 – 62.92960 = 0.59194 u Binding energy per nucleus: EB = 0.59194 × 931.5 = 551.4 MeV Total energy for all nuclei: E_total = N × EB = 2.868 × 10²² × 551.4 MeV = 1.581 × 10²⁵ MeV = 1.581 × 10²⁵ × 1.6 × 10⁻¹³ J = 2.53 × 10¹² J

Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and the silver isotope ¹⁰⁷₄₇Ag.

Solution

Nuclear radius: R = R₀ × A^(1/3) R₀ = 1.2 × 10⁻¹⁵ m (same for all nuclei) Ratio of radii: R_Au/R_Ag = (A_Au/A_Ag)^(1/3) = (197/107)^(1/3) = (1.8411)^(1/3) = 1.2255 ≈ 1.23 R_Au : R_Ag ≈ 1.23 : 1

Find the Q-value and the kinetic energy of the emitted alpha particle in the alpha decay of (a) ²²⁶Ra and (b) ²²⁰Rn. Given: m(²²⁶Ra) = 226.02540 u, m(²²²Rn) = 222.01750 u, m(²²⁰Rn) = 220.01137 u, m(²¹⁶Po) = 216.00189 u.

Solution

m_α = 4.00260 u (a) ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He Q = [m(Ra) – m(Rn) – m(α)] × 931.5 MeV = [226.02540 – 222.01750 – 4.00260] × 931.5 = [0.00530] × 931.5 = 4.94 MeV KE of alpha particle: KE_α = Q × (A – 4)/A = 4.94 × 222/226 = 4.85 MeV (b) ²²⁰₈₆Rn → ²¹⁶₈₄Po + ⁴₂He Q = [220.01137 – 216.00189 – 4.00260] × 931.5 = [0.00688] × 931.5 = 6.41 MeV KE_α = 6.41 × 216/220 = 6.29 MeV

The radionuclide ¹¹C decays according to: ¹¹₆C → ¹¹₅B + e⁺ + ν. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m(¹¹₆C) = 11.011434 u and m(¹¹₅B) = 11.009305 u. Calculate Q and compare it with the maximum energy of the positron emitted.

Solution

For β⁺ decay: Q = [m_N(¹¹C) – m_N(¹¹B) – 2m_e] × 931.5 MeV Using atomic masses: Q = [M(¹¹C) – M(¹¹B) – 2m_e] × 931.5 m_e = 0.000549 u Q = [11.011434 – 11.009305 – 2 × 0.000549] × 931.5 = [11.011434 – 11.009305 – 0.001098] × 931.5 = [0.001031] × 931.5 = 0.9603 MeV ≈ 0.960 MeV The Q value (0.960 MeV) equals the maximum kinetic energy of the emitted positron ✓ (The maximum KE of positron = Q when the neutrino carries no energy)

Q10

The Q value of a nuclear reaction A + b → C + d is defined by Q = (mA + mb – mC – md)c². If the Q value of a reaction is positive, the reaction is said to be exothermic. Consider the reaction ¹H + ³H → ²H + ²H. Is this exothermic or endothermic? m(¹H) = 1.007825 u, m(²H) = 2.014102 u, m(³H) = 3.016049 u.

Solution

Reaction: ¹H + ³H → ²H + ²H Q = [m(¹H) + m(³H) – 2m(²H)] × 931.5 MeV = [1.007825 + 3.016049 – 2 × 2.014102] × 931.5 = [4.023874 – 4.028204] × 931.5 = [–0.004330] × 931.5 = –4.03 MeV Since Q < 0, the reaction is endothermic. Energy of 4.03 MeV must be supplied for this reaction to occur.

Q11

Suppose, we think of fission of a ⁵⁶Fe nucleus into two equal fragments of ²⁸Al. Is the fission energetically possible? Argue by working out Q of the process. m(⁵⁶Fe) = 55.93494 u and m(²⁸Al) = 27.98191 u.

Solution

⁵⁶Fe → ²⁸Al + ²⁸Al Q = [m(⁵⁶Fe) – 2m(²⁸Al)] × 931.5 = [55.93494 – 2 × 27.98191] × 931.5 = [55.93494 – 55.96382] × 931.5 = [–0.02888] × 931.5 = –26.9 MeV Since Q < 0, the fission is energetically NOT possible. Energy would need to be supplied. This makes sense: ⁵⁶Fe is near the peak of the binding energy per nucleon curve (~8.8 MeV/nucleon), while ²⁸Al has lower binding energy per nucleon. You cannot get energy by splitting ⁵⁶Fe.

Q12

The fission properties of ²³⁹Pu are very similar to those of ²³⁵U. The average energy released per fission of ²³⁹Pu is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ²³⁹Pu undergo fission?

Solution

Mass = 1 kg = 1000 g, M(²³⁹Pu) = 239 g/mol Number of atoms: N = (1000/239) × 6.022 × 10²³ = 4.182 × 0.6022 × 10²⁴ ... wait: N = (1000/239) × 6.022 × 10²³ = 4.182 × 6.022 × 10²³? No: N = (1000/239) × 6.022 × 10²³ = 4.1841 × 6.022 × 10²³ Actually: 1000/239 = 4.184 N = 4.184 × 6.022 × 10²³ = 25.2 × 10²³ = 2.52 × 10²⁴ Energy per fission = 180 MeV Total energy = N × 180 = 2.52 × 10²⁴ × 180 = 4.536 × 10²⁶ MeV ≈ 4.54 × 10²⁶ MeV

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