Consider the neutron–induced fission reaction _0^ 1 n+_ 92 ^ 235 U ; ;_ 56 ^ 141 Ba…

PHYSICS · Subjective · CBSE · Physics · Nuclei · Nuclei

Problem

Consider the neutron–induced fission reaction 01n+92235U  ⟶  56141Ba+3692Kr+3 01n_0^{1}\text n+_{92}^{235}\text U\;\longrightarrow\;_{56}^{141}\text{Ba}+_{36}^{92}\text{Kr}+3\,_0^{1}\text n The atomic masses are: m(92235U)=235.043930 um(_{92}^{235}\text U)=235.043930\,\text u, m(01n)=1.008665 um(_0^{1}\text n)=1.008665\,\text u, m(56141Ba)=140.914411 um(_{56}^{141}\text{Ba})=140.914411\,\text u, m(3692Kr)=91.926156 um(_{36}^{92}\text{Kr})=91.926156\,\text u. Take 1 u=931.5 MeV/c21\,\text u = 931.5\,\text{MeV}/c^{2}, NA=6.022×1023 mol−1N_{A}=6.022\times10^{23}\,\text{mol}^{-1}, and 1 MeV=1.602×10−13 J1\,\text{MeV}=1.602\times10^{-13}\,\text J. (a) Calculate the Q–value of the fission reaction in MeV. (b) Estimate the energy released per mole of  92235U\,_{92}^{235}\text U that undergoes complete fission. (c) If an ideal reactor could cause complete fission of 1 kg1\,\text{kg} of  92235U\,_{92}^{235}\text U, compute the total energy liberated in joules. (d) The combustion of 1 kg1\,\text{kg} of coal yields roughly 3×107 J3\times10^{7}\,\text J. Calculate how many kilograms of coal would be required to produce the same amount of energy as in part (c). Comment briefly on the significance of your result.

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