Question

Nuclear reactor in which $U-235$ is used as fuel. Uses $2\, kg$ of $U-235$ in $30$ days. Then, power output of the reactor will be (given energy released per fission $=185\, M eV$ )

• A. 43.5 MW
• B. 58.5 MW
• C. 69.6 MW
• D. 73.1 MW

Answer 1

Answer: (B)

Number of atoms in $2\, kg$ of uranium
$=\frac{6.02 \times 10^{23}}{235} \times 2000$
$=5.12 \times 10^{24}$
Therefore, energy obtained from these atoms
$=5.12 \times 10^{24} \times 185\, MeV$
$=5.12 \times 10^{24} \times 185 \times 10^{6} eV$
Energy obtained per second $=\frac{5.12 \times 10^{24} \times 185 \times 10^{6} \times 16 \times 10^{-19}}{30 \times 24 \times 60 \times 60}$
Solving the above expression $=5847 \times 10^{6} W \cong 58.5\, MW$