DHS 20 Nuclear Physics (Tutorial Solutions)

Dunman High School (Senior High Y6 Physics) 2025 Tutorial Solutions: Nuclear Physics For Internal Use Only Page 1 of 9 No Solution Ans/Mark 1 The reason that most of the α-particles are either undeflected or deflected through very small angles is because the nucleus is much smaller than the atom. A few α-particles were deflected through angles as large as 90o or more. Only if the positive charge of the atom1 is concentrated in a more compact region that a much larger force will occur at near impacts. The large-angle scattering obtained experimentally could result only from a single encounter of the α-particle with a massive charge confined to a volume much smaller than that of the whole atom. [D] 2 A neutron has no charge, thus will not be affected by electric or magnetic fields. The mass of a neutron is the same as that of a proton. Can you identify the other 3 particles? [D] 3 According to the conservation of mass number and atomic number, Mass number of bohrium, x = 255 + 3(4) = 267, i.e., 267 nucleons Also, proton number of bohrium, y = 101 + 3(2) = 107, i.e., 107 protons. [A] 4 (a) F = 12 2 04 QQ xπε = = . 1 1 (b) (i) Loss in K.E. = gain in P.E. 1.8 MeV = (x) x = = 1.26×10−13 m. 1 1 (ii) Radii of the two nuclei are less than 1.26×10 −13 m, : What are the approximate radii of the two nuclei? 1 1 If the atom consisted of a positively charged sphere of radius 10−10 m, containing electrons as in the Thomson model, only a very small deflection could result from a single encounter between an α-particle and an atom, even if the α - particle penetrated into the atom. The Thomson atomic model could not possibly account for the number of large-angle scatterings that Rutherford saw. 2 (2 )(79 ) 4 o ee xπε 26 2 3.64 10 x −× 26 2 3.64 10 x −× 26 13 3.64 10 1.8(1.6 10 ) − − × × x α (He nucleus) Au nucleus x = 255 + 3 (4) y = 101 + 3 (2) 4 2 4 2 4 2

Dunman High School (Senior High Y6 Physics) 2025 Tutorial Solutions: Nuclear Physics For Internal Use Only Page 2 of 9 5 The graph of binding energy per nucleon BE against nucleon number shows that, with increasing nucleon number, BE increases steeply initially but reaches a peak at 56Fe and decreases gradually beyond that. This shows that A, B and D are false. Energy will be released if a massive nucleus with low B E is broken up, as a result of particle bombardment, into two or more lighter nuclei, each with greater BE. The daughter nuclei are more stable. Ans: [C] 6 Since 0.001 u has an energy equivalent of 0.9 MeV and α particle has to travel and collide with nitrogen, thus the kinetic energy of the reactants exceeds the kinetic energy of the products by 0.9 MeV. Ans: [C] Option D seems possible but it cannot be the correct answer. Why? [ Consider using the principle of conservation of linear momentum.] 7 (a) Energy released = ∆m.c2 = [ 3.753 – (3.686+0.066)] ×10−25 kg (3×108 m s−1)2 = 9.0×10−12 J. 1 1 (b) E = hf = hc λ So, λ = hc E = 34 8 1 12 (6.63 10 J s)(3 10 m s ) (9.0 10 J)(4 100) −− − ×× × = 5.53×10−13 m. 1 1 (c) The remaining 96% of the energy is in the form of K.E. of 222Rn and 4He . 1 8 (a) Loss of mass = (4.00260 u + 9.01212 u) − (1.00867 u + 12.00000 u) = 6.05 × 10−3 u 1 1 (b) Energy equivalence of this mass = mc2 = (6.05 × 10−3 u) c2 = (6.05 × 10−3 × 1.66 × 10−27)( 3.00 × 108 )2 = 9.04 × 10−13 J 1 1 9 (a) 17 8 O nucleus comprises 8 protons and 9 neutrons. 1,1 (b) + →+14 4 17 1 72 81N He O H Change in mass, ∆m = 18.00696 u - 18.00568 u = 0.00128 u = (0.00128)(1.66×10−27 kg) = 2.125×10−30 kg. The minimum K.E. of the α-particle is the energy equivalent of this ∆m, i.e. E = ∆m c2 = (2.125×10−30 kg)(3×108 m s−1)2 = 1.91×10−13 J. 1 1 1 1

Dunman High School (Senior High Y6 Physics) 2025 Tutorial Solutions: Nuclear Physics For Internal Use Only Page 3 of 9 (c) Taking to the right as positive, ( + →+14 4 17 1 72 81N He O H ) (mv)α = (mv)o + (mv)p (4 u)(3.0×107 m s−1) = (17 u) vo + (1 u)(6.0×107 m s−1) ⇒ vo = 3.53×106 m s−1 --- case (1) same direction as the initial velocity of the alpha particle Note: If the proton moves in the opposite direction of the initial velocity of the alpha particle, then (4 u)(3.0×107 m s−1) = (17 u) vo − (1 u)(6.0×107 m s−1) ⇒ vo = 1.06×107 m s−1 --- case (2) Using conservation of mass-energy, total final kinetic energy of system = (total initial kinetic energy of system) – (energy equivalent of the increase in mass) = [½ (4 u) (3.0×107)2] – (1.91×10−13) = 2.8 × 10−12 J ----- (*) For case (1), total final kinetic energy of system = ½ (17 u) (3.53×106)2 + ½ (1 u) (6.0×107)2 = 3.2 × 10−12 J [very close to the value in (*); they are equal, to 1 s.f.] For case (2), total final kinetic energy of system = ½ (17 u) (1.06×107)2 + ½ (1 u) (6.0×107)2 = 4.6 × 10−12 J [larger than the value in (*) – defy conservation of mass-energy!] Thus, the oxygen-17 and the proton move in the same direction as that of the initial velocity of the alpha particle, after the nuclear reaction, and the velocity of the oxygen-17 nucleus is 3.53×106 m s−1. 1 1 1

Dunman High School (Senior High Y6 Physics) 2025 Tutorial Solutions: Nuclear Physics For Internal Use Only Page 4 of 9 10 The total mass of the products of this process is equivalent to 939 MeV + 940 MeV = 1879 MeV. The mass of the original deuteron is equivalent to 1876 MeV For the deuteron to disintegrate to a proton and a neutron, there is a mass defect equivalent to 1879 MeV – 1876 MeV = 3 MeV Therefore, the deuteron needs to capture a γ-ray photon of energy 3 MeV. This 3 MeV is the binding energy of the deuteron, which is the minimum energy required for the deuteron to break into a proton and a neutron. [D] 11 The process may be represented by 27 13 Al  4 2 He + 23 11 Na The total mass of the products = 4.0026 u + 22.9898 u = 26.9924 u This is more than the mass of 27 13 Al (which is 26.9815 u) by 26.9924 u – 26.9815 u = 0.0109 u Mass-energy equivalence: E = mc2: E = (0.0109 u) × c2 = 0.0109 × 1.66 × 10−27 × (3 × 108)2 = 1.63 × 10−12 J Thus 0.0109 u is equivalent to 1.63 × 10−12 J of energy. So 1.63×10−12 J of energy is required for this process to occur, and it would not occur spontaneously. Note - The balanced nuclear equation for the induced transmutation of Aluminum-27 into Sodium-24 by neutron bombardment, with release of an alpha particle in the reaction: 1 0 27 4 24 13 2 11Al He Na+→ +n 1 1 1 1 12 current = rate of flow of charges 0.01×10−6 = 191 10 where is the number of ions per secondNN tt −×× N t⇒ = 1 × 1011 ⇒ Average number of ions produced by each α-particle = 11 6 1 10 1 10 × × = 1 × 105 [A] 13 γ – radiation is chosen over β– particles because of its relatively better penetrating power so that it is not easily blocked through the 0.40 m of solid obstacles beneath the field. The half-life of the radioactive material needs to be small so that its activity will fall to a biologically safe level quic