Question

A nuclear power reactor produces 3GW of thermal power from the fission of 235U for a period of 10 weeks and then it is shut down. Note: All numerical answers are expected with 3 significant figures. (a) Assuming 78% of the 200MeV released per fission event is converted to heat, estimate the number of 235U nuclei fissioned over the 10 week period. (b) 6 % of the 235U fission events produce as a fission fragment some radioactive nuclide X (e.g. 137Cs), which has a half-life of 38 years. Consider the amount of X nuclei produced during the 10 week operation of the reactor in Part A and calculate the activity of X left in the used fuel 38 years after the reactor is stopped. (c) Following Part B, estimate the power (in kW) produced by these decays of X 38 years after the reactor shutdown if each decay of X releases 1.6 MeV .

Answer 1

a) Total energy produced in 10 weeks = (3*109)*(10*7*24*60*60) = 1.8144 * 1016 J

Energy per fission = 200 MeV

Energy converted to heat per fission = 0.78*200 = 156 MeV = (156*106)*(1.6*10-19) J = 2.496*10-11 J

Hence, total number of fissions = (1.8144 * 1016 J)/(2.496*10-11 J) = 7.27*1026 ....Answer

b) Number of X formed during 10 weeks = 0.06*7.27*1026 = 4.362*1025

Since, t0.5 = ln2/,   =ln2/t0.5 = 0.693/(38 years) = 0.01824 year-1

Now, 10 weeks << 38 years. Hence, we can roughly assume that almost no nuclei of X decay during the initial 10 weeks. After 38 years, half of X would have decayed.

Hence, number of undecayed X after 38 years = (4.362*1025)/2 = 2.181*1025

Activity = X = (0.01824 year-1) * 2.181*1025 = 3.977*1023/year = 3.977*1023/(365*24*60*60)/second = 1.26*1016 dps .....Answer

c) Each decay produces 1.6 MeV.

Decay rate after 38 years = 1.26*1016 dps

Hence power produced = 1.6 MeV*1.26*1016 dps = (1.6*103)*(1.6*10-19)*(1.26*1016) = 3.23 kW ....Answer