 # Solve this following

Question:

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

1. $125 \mathrm{MW}$

2. $60 \mathrm{MW}$

3. $35 \mathrm{MW}$

4. $54 \mathrm{MW}$

Correct Option: , 2

Solution:

Number of uranium atoms in $2 \mathrm{~kg}$

$=\frac{2 \times 6.023 \times 10^{26}}{235}$

energy from one atom is $200 \times 10^{6}$ e.v. hence total energy from $2 \mathrm{~kg}$ uranium

$=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$

$2 \mathrm{~kg}$ 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