Physics

Physics

Problem

A typical fission process of uranium-235 may be represented, in a simplified form, by   92235U+01n→  56141Ba+  3692Kr+3 01n^{235}_{\;92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{141}_{\;56}\text{Ba} + ^{92}_{\;36}\text{Kr} + 3\,^{1}_{0}\text{n} Atomic masses: 235U=235.043929 u^{235}\text{U} = 235.043929\,\text{u}, 141Ba=140.914411 u^{141}\text{Ba} = 140.914411\,\text{u}, 92Kr=91.926156 u^{92}\text{Kr} = 91.926156\,\text{u}, 01n=1.008665 u^{1}_{0}\text{n} = 1.008665\,\text{u}. Take 1 u=931.5 MeV/c21\,\text{u} = 931.5\,\text{MeV}/c^2. a. (i) Write the mass defect Δm\Delta m for one fission event, showing all terms clearly. b. (ii) Calculate the energy released per fission event in MeV. c. (iii) Determine the total energy obtainable (in joule) if 1.0 kg1.0\,\text{kg} of 235U^{235}\text{U} undergoes complete fission. (Avogadro number NA=6.02×1023 mol−1N_A = 6.02\times10^{23}\,\text{mol}^{-1}.) d. (iv) Coal releases roughly 3.0×107 J3.0\times10^{7}\,\text{J} per kilogram when burnt. Estimate how many kilograms of coal would be needed to release the same amount of energy as in part (iii).

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