Question

In a reactor, $2 kg$ of $_{92} U ^{235}$ fuel is fully used up in 30 days. The energy released per fission is $200 MeV$. Given that the Avogadro number, $N =6.023 \times 10^{26}$ per kilo mole and $1 eV =1.6 \times 10^{-19} J$. The power output of the reactor is close to :

• A. $125 MW$
• B. $60 MW$
• C. $35 MW$
• D. 54 MW

Answer 1

Answer: (B)

Number of uranium atoms in $2 kg$
$=\frac{2 \times 6.023 \times 10^{26}}{235}$
energy from one atom is $200 \times 10^{6}$ e.v. hence total energy from $2 kg$ uranium
$=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} J$
$2 kg$ uranium is used in 30 days hence this energy is recieved in 30 days hence energy recived per second or power is
Power $=\frac{2 \times 6.023 \times 10^{26} \times 200 \times 10^{6} \times 1.6 \times 10^{-19}}{235 \times 30 \times 24 \times 3600}$
Power $=63.2 \times 10^{6}$ watt or 63.2 Mega Watt