CAIE A-Level Physics 23.1.7 Energy Released in Nuclear ReactionsPractise calculating nuclear reaction energy from mass change or binding-energy change, using E = c² delta m and converting joules and electronvolts.Syllabus2028–2030CoursePhysics 9702LevelA2
Exam pointscalculate energy released in a nuclear reaction from mass change using E = c²Δmcalculate energy released from changes in total binding energy and convert consistently between joules and electronvolts
23.1.7—The energy released in nuclear reactions using E = c 2∆m question 1[Maximum number: 5]Table 8.1 shows the masses of the particles involved in the reaction in (b).Table 8.1Calculate the energy released when 1.00 mol of deuterium undergoes the reaction. JShow AnswerMark as masteredΔm=[(2×2.014102)−(3.016029+1.008665)]u(=0.00351u)\begin{aligned} \Delta m =[(2 \times 2.014102)-(3.016029+1.008665)] \mathrm{u} (=0.00351 \mathrm{u}) \end{aligned}Δm=[(2×2.014102)−(3.016029+1.008665)]u(=0.00351u)C1E=Δmc2E=\Delta m c^{2}E=Δmc2C1=0.00351×1.66×10−27×(3.00×108)2(=5.24×10−13 J)\begin{aligned} =0.00351 \times 1.66 \times 10^{-27} \times\left(3.00 \times 10^{8}\right)^{2} \left(=5.24 \times 10^{-13} \mathrm{~J}\right) \end{aligned}=0.00351×1.66×10−27×(3.00×108)2(=5.24×10−13 J)C11.00 mol of deuterium forms 0.500 mol of helium-3C1 total energy =0.500×6.02×1023×5.24×10−13=1.58×1011 J\begin{aligned} \text { total energy } =0.500 \times 6.02 \times 10^{23} \times 5.24 \times 10^{-13} =1.58 \times 10^{11} \mathrm{~J} \end{aligned} total energy =0.500×6.02×1023×5.24×10−13=1.58×1011 JA1Add to Test
