A reasonable estimate is that 100 times 109 ton of natural u

A reasonable estimate is that 1.00 times 10^9 ton of natural uranium is available at concentrations exceeding 100 parts per million, of which 0.7% is the fissionable isotope 235U. Assume all the world\'s energy use (7.0 times 10^12 J/s) is supplied by 235 U fission in conventional nuclear reactors, releasing 208 MeV for each reaction. How long would the supply last? (The ton used in this exercise is a metric ton, or 1000 kg.)

Solution

Mass of Uranium, m = 1 x 10^9 x 10^3 kg x 0.7 / 100

       = 7 x 10^9 kg

mass of 1 mol 235 U = 235 grams

moles of uranium = (7 x 10^9 x 10^3 grams) / (235 grams/mole)

       = 29.787 x 10^9 moles

number of nucleaus = 29.787 x 10^9 x 6.022 x 10^23 = 1.794 x 10^34

energy by fission = 1.794 x 10^34 x 208 MeV

    = 1.794 x 10^34 x 208 x 10^6 x 1.6 x 10^-19 J

= 5.97 x 10^23 J

time for energy will be sufficient, t = (5.97 x 10^23) / (7 x 10^12)

        = 8.528 x 10^10 s

t = (8.528 x 10^10 ) / ( 3600 x 24 x 365 )   = 2704.27 years