Consider the neutron–induced fission reaction
01n+92235U⟶56141Ba+3692Kr+301n
The atomic masses are: m(92235U)=235.043930u, m(01n)=1.008665u, m(56141Ba)=140.914411u, m(3692Kr)=91.926156u. Take 1u=931.5MeV/c2, NA=6.022×1023mol−1, and 1MeV=1.602×10−13J.
(a) Calculate the Q–value of the fission reaction in MeV.
(b) Estimate the energy released per mole of 92235U that undergoes complete fission.
(c) If an ideal reactor could cause complete fission of 1kg of 92235U, compute the total energy liberated in joules.
(d) The combustion of 1kg of coal yields roughly 3×107J. 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.