WJEC Physics for A2: Study and Rev Guide

224 A Level Physics: Study and Revision Guide Q & A 6 (a) Calculate the binding energy per nucleon of 14 6 C . [4] [ 1u º 931 MeV, m neutron =1.008665 u , m proton =1.007276 u , mass of 14 6 C nucleus = 13.999950u ] The following reaction can be regarded as evidence for the existence of neutrinos (or an anti-neutrino in this case). 14 6 C ¾¾® 14 7 N + 0 −1 b + – ν e Mass of 14 6 C nucleus = 13.999950u ; Mass of 14 7 N nucleus = 13.999234u Mass of b − particle = 0.000549u ; The mass of the anti-neutrino – ν e is negligible. (b) Calculate the energy released in this reaction. [3] The evidence for the existence of the anti-neutrino came from the (unexpected) wide variation of the energies of the b -particles emitted. However, you should now ignore the existence of the anti-neutrino. (c) Explain briefly, using conservation of momentum, which particle ( N or b – ) receives most of the energy of the reaction. [3] 14 6 C 14 7 N Before the reaction (stationary ) β − After the reaction 14 6 C Tom’s answer (a) m total = 6 ´ 1.007276 + 8 ´ 1.008665 ✓ = 14.112976 u m lost = 14.112976 – 13.999950 = 0.113026 u ✓ E = mc 2 = 0.113026 ´ (3 ´ 10 8 ) 2 = 1.017234 ´ 10 16 eV ✗ (b) 13.999950 – (13.999234 + 0.000549) = 1.67 ´ 10 - 4 u ✓ E = mc 2 = 1.67 ´ 10 - 4 ´ (3 ´ 10 8 ) 2 = 1.503 ´ 10 13 eV (c) The nitrogen nucleus receives most of the energy from the reaction because it is a lot heavier than the b / electron emitted ✓ and the 14 7 N still has a fair bit of velocity. \ the 14 7 N has the most energy from the collision. Examiner commentary (a) Tom has calculated the mass of the 14 nucleons correctly and the mass deficit of the nucleus in u – giving him the first two marks. He wants to use E = mc 2 , so he should have converted the u to kg . He also fails to divide by the number of nucleons. (b) Tom’s first step is correct but he makes the same mistake as in part (a). (c) Tom correctly identifies the significant fact of the relative masses of the 14 7 N and the b -particle but fails to draw the correct conclusion. Tom scores 4 out of 10 marks. Seren’s answer (a) 6 ´ 1.007276 + 8 ´ 1.008665 ✓ = 14.112976 u 14.112976 – 13.999950 = 0.113026 u ✓ × 0.113026 931 14 ✓ = 7.52 MeV ✓ (b) 13.999950 – (13.999234 + 0.000549) = 1.67 ´ 10 - 4 u ✓ Energy released = 931 ´ 1.67 ´ 10 - 4 ✓ = 0.155 MeV ✓ (c) The momentums of the nucleus and the b particle, ie mv for the two particles are the same and so v for the lighter b particle ✓ is much higher ✓ . Kinetic energy is 1 2 mv 2 which is 1 2 mv ´ v , so the b particle with the bigger velocity has the bigger KE ✓ . Examiner commentary (a) Seren has done all the steps and has received the marks. Her communication is not perfect because she doesn’t say what each line of the answer means. Her answer is correct [the units MeV, MeV/nucleon were accepted]. (b) Again correct with better communication. (c) Seren has come to the correct conclusion with good reasoning. Even easier would be to note that the relationship between KE and momentum is KE = p 2 2 m , so if the momenta are the same, the lighter particle gets the lion’s share of the energy. Seren scores 10 out of 10 marks.