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 see