Data given in the book: e = 1.6 × 10⁻¹⁹ C, N_A = 6.023 × 10²³ per mole, 1/(4πε₀) = 9 × 10⁹ N m²/C², 1 MeV = 1.6 × 10⁻¹³ J, 1 u = 931.5 MeV/c², 1 year = 3.154 × 10⁷ s, m_H = 1.007825 u, mₙ = 1.008665 u, m(⁴₂He) = 4.002603 u, mₑ = 0.000548 u.
Exercises
(13.1) Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given m(¹⁴₇N) = 14.00307 u.
Answer
¹⁴₇N has Z = 7 protons and N = 7 neutrons. Using atomic masses (electron masses cancel):
Mass defect Δm = 7m_H + 7mₙ − m(N) = 7 × 1.007825 + 7 × 1.008665 − 14.00307
⇒ Δm = 7.054775 + 7.060655 − 14.00307 = 0.11236 u
E_b = Δm × 931.5 MeV = 0.11236 × 931.5
⇒ E_b ≈ 104.66 MeV
(13.2) Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi in units of MeV from the following data: m(⁵⁶₂₆Fe) = 55.934939 u, m(²⁰⁹₈₃Bi) = 208.980388 u.
Answer
⁵⁶₂₆Fe: Z = 26, N = 30.
Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.20345 + 30.25995 − 55.934939
⇒ Δm = 0.528461 u
E_b = 0.528461 × 931.5
⇒ E_b ≈ 492.26 MeV (per nucleon ≈ 492.26/56 ≈ 8.79 MeV)
²⁰⁹₈₃Bi: Z = 83, N = 126.
Δm = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.09179 − 208.980388
⇒ Δm = 1.760877 u
E_b = 1.760877 × 931.5
⇒ E_b ≈ 1640.26 MeV (per nucleon ≈ 1640.26/209 ≈ 7.85 MeV)
So iron has the larger binding energy per nucleon and is more stable than bismuth.
(13.3) A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ⁶³₂₉Cu atoms (of mass 62.92960 u).
Answer
Number of atoms in 3.0 g: N = (3.0/62.9296) × 6.023 × 10²³
⇒ N ≈ 2.87 × 10²² atoms
⁶³₂₉Cu has Z = 29, N = 34. Δm = 29 × 1.007825 + 34 × 1.008665 − 62.92960 = 29.226925 + 34.29461 − 62.92960
⇒ Δm = 0.591935 u
Binding energy per nucleus = 0.591935 × 931.5 ≈ 551.4 MeV
Total energy = 2.87 × 10²² × 551.4 MeV
⇒ ≈ 1.58 × 10²⁵ MeV = 1.58 × 10²⁵ × 1.6 × 10⁻¹³ J
⇒ ≈ 2.53 × 10¹² J
(13.4) Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and the silver isotope ¹⁰⁷₄₇Ag.
Answer
R = R₀A^(1/3)
⇒ R_Au/R_Ag = (197/107)^(1/3) = (1.841)^(1/3)
⇒ R_Au/R_Ag ≈ 1.23
(13.5) The Q value of a nuclear reaction A + b → C + d is defined by Q = [m_A + m_b − m_C − m_d]c², where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) ¹₁H + ³₁H → ²₁H + ²₁H (ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He. Atomic masses are given to be m(²₁H) = 2.014102 u, m(³₁H) = 3.016049 u, m(¹²₆C) = 12.000000 u, m(²⁰₁₀Ne) = 19.992439 u.
Answer
Atomic masses can be used directly because the number of electrons is the same on both sides, so electron masses cancel. Take m(¹₁H) = 1.007825 u and m(⁴₂He) = 4.002603 u.
(i) Δm = (1.007825 + 3.016049) − 2 × 2.014102 = 4.023874 − 4.028204
⇒ Δm = −0.00433 u
Q = −0.00433 × 931.5
⇒ Q ≈ −4.03 MeV. Q is negative, so the reaction is endothermic.
(ii) Δm = 2 × 12.000000 − (19.992439 + 4.002603) = 24.000000 − 23.995042
⇒ Δm = 0.004958 u
Q = 0.004958 × 931.5
⇒ Q ≈ +4.62 MeV. Q is positive, so the reaction is exothermic.
(13.6) Suppose, we think of fission of a ⁵⁶₂₆Fe nucleus into two equal fragments, ²⁸₁₃Al. Is the fission energetically possible? Argue by working out Q of the process. Given m(⁵⁶₂₆Fe) = 55.93494 u and m(²⁸₁₃Al) = 27.98191 u.
Answer
Process: ⁵⁶₂₆Fe → ²⁸₁₃Al + ²⁸₁₃Al
Q = [m(Fe) − 2m(Al)]c² = [55.93494 − 2 × 27.98191] × 931.5 MeV = (−0.02888) × 931.5
⇒ Q ≈ −26.9 MeV
Q is negative, so energy would have to be supplied; the fission is not energetically possible on its own.
(13.7) The fission properties of ²³⁹₉₄Pu are very similar to those of ²³⁵₉₂U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ²³⁹₉₄Pu undergo fission?
Answer
Number of atoms in 1 kg: N = (1000/239) × 6.023 × 10²³
⇒ N ≈ 2.52 × 10²⁴
Energy released = N × 180 MeV = 2.52 × 10²⁴ × 180
⇒ E ≈ 4.54 × 10²⁶ MeV
(13.8) How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as ²₁H + ²₁H → ³₂He + n + 3.27 MeV.
Answer
Number of deuterium nuclei in 2.0 kg (molar mass ≈ 2 g/mol): N = (2000/2) × 6.023 × 10²³
⇒ N ≈ 6.02 × 10²⁶
Each fusion uses 2 nuclei, so number of reactions = 3.01 × 10²⁶
Energy released E = 3.01 × 10²⁶ × 3.27 MeV ≈ 9.85 × 10²⁶ MeV
⇒ E = 9.85 × 10²⁶ × 1.6 × 10⁻¹³ J ≈ 1.58 × 10¹⁴ J
Time t = E/P = 1.58 × 10¹⁴/100 = 1.58 × 10¹² s
⇒ t = 1.58 × 10¹²/3.154 × 10⁷ years ≈ 5.0 × 10⁴ years (about 4.9 × 10⁴ years if the molar mass 2.014 g is used)
(13.9) Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Answer
When the deuterons just touch, the distance between their centres is r = 2 × 2.0 fm = 4.0 × 10⁻¹⁵ m. Each has charge e.
U = (1/4πε₀)e²/r = 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/(4.0 × 10⁻¹⁵)
⇒ U = 5.76 × 10⁻¹⁴ J
U = 5.76 × 10⁻¹⁴/1.6 × 10⁻¹⁹ eV
⇒ U ≈ 3.6 × 10⁵ eV = 360 keV
So the height of the Coulomb barrier is about 360 keV.
(13.10) From the relation R = R₀A^(1/3), where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
Answer
Mass of a nucleus M ≈ A × m, where m is the mass of one nucleon (≈ 1.67 × 10⁻²⁷ kg).
Volume V = (4/3)πR³ = (4/3)π(R₀A^(1/3))³
⇒ V = (4/3)πR₀³A
Density ρ = M/V = Am/[(4/3)πR₀³A]
⇒ ρ = 3m/(4πR₀³)
A cancels, and m and R₀ are constants, so the nuclear density is the same for all nuclei (about 2.3 × 10¹⁷ kg/m³ taking R₀ = 1.2 fm).
Important Extra Questions
What are isotopes, isobars and isotones? Give one example of each.
Answer
Isotopes: same Z, different A (¹²₆C, ¹⁴₆C). Isobars: same A, different Z (³₁H, ³₂He). Isotones: same neutron number N, different Z (¹⁹⁸₈₀Hg, ¹⁹⁷₇₉Au).
Why is the binding energy per nucleon lower for very light and very heavy nuclei?
Answer
In light nuclei each nucleon has few neighbours to bond with (more surface nucleons). In heavy nuclei the Coulomb repulsion among many protons grows and reduces binding, while the short-range nuclear force does not increase. So E_bn peaks (≈ 8.75 MeV) near A = 56.
State three properties of the nuclear force.
Answer
It is the strongest force, much stronger than the Coulomb force; it is short-ranged (a few femtometres), becoming repulsive below about 0.8 fm; and it is nearly the same for n–n, n–p and p–p pairs (charge independent).
Why is a very high temperature needed for nuclear fusion?
Answer
The positively charged nuclei must have enough kinetic energy to overcome their Coulomb repulsion (a barrier of hundreds of keV) and come within the range of the nuclear force; such energies occur only at temperatures of the order of 10⁷–10⁹ K.
Express 1 u in MeV.
Answer
E = mc² = 1.6605 × 10⁻²⁷ × (3 × 10⁸)² J ≈ 1.49 × 10⁻¹⁰ J = 1.49 × 10⁻¹⁰/1.6 × 10⁻¹³ MeV ≈ 931.5 MeV.