Question

Assuming that about $200MeV$ 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

Answer: 1.00

$200MeV=200\times 1.6\times 10^{-13}J$
$=3.2\times 10^{-11}J/$ 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.