Question

Calculate power output of ${}_{92}^{235}U$ reactor, if it takes $30days$ to use up $2kg$ of fuel, and if each fission gives $185MeV$ of useable energy. Avogadro's number $=6\times {10}^{23}/mol$?

• A. $56.3MW$
• B. $60.3MW$
• C. $58.3MW$
• D. $54.3MW$

Answer 1

The correct option is C $58.3MW$

No. of ${}^{235}U$ atoms in $2kg$$ of fuel
$=\frac{6.023\times {10}^{23}}{235}\times 2000$
fission energy per atom $=185MeV$
$\therefore$ Energy for $2kg$ of fuel
$=\frac{6.023\times {10}^{26}\times 2}{235}\times 185MeV$
$Power=\frac{\text{Energy released}}{time}$
$=\frac{6.023\times {10}^{26}\times 2\times 185\times 1.6\times {10}^{-13}J}{235\times 30days}$
$(\because 1MeV=1.6\times {10}^{-13}J,30days=30\times 24\times 60\times 60sec)$
$\therefore Power=\frac{6.023\times {10}^{26}\times 2\times 185\times 1.6\times {10}^{-13}}{235\times 30\times 24\times 60\times 60}$
$=\frac{3552\times {10}^{13}}{235\times 3\times 6\times 6\times 24\times {10}^3}W$
$=\frac{3552\times {10}^{10}}{235\times 3\times 6\times 6\times 24}$
$=\frac{3552\times {10}^4}{235\times 18\times 6\times 24}MW=58.3MW$