Question

Assuming that about $200 \,MeV$ of energy is released per fission of ${ }_{92} U ^{235}$ nuclei, then the mass of $U^{235}$ consumed per day in a fission reactor of power $1$ megawatt will approximately be (in grams)

Answer 1

$200 MeV =200 \times 1.6 \times 10^{-13}\, J$
$=3.2 \times 10^{-11} J /$ fission
or $3.2 \times 10^{-11} J /$ Atom
Power required $=1$ Mega Watt $=10^{6} J / s$
No. of atoms required per second
$=\frac{10^{6}}{3.2 \times 10^{-11}}$
$=3.125 \times 10^{16}$ atoms/sec.
no. of atoms per day
$=3.125 \times 10^{6} \times 8.64 \times 10^{4}$
$=2.7 \times 10^{21}$ atoms/day
Mass consumed per day
$=\frac{235}{6 \times 10^{23}} \times \frac{2.7 \times 10^{21}}{1}$
$=1.05$ gram.