Question

A ${U^{235}}$ nuclear reactor generates energy at a rate of $3.70 \times {10^7}J/s$. Each fission liberates 185MeV useful energy. If the reactor has to operate for $144 \times {10^4}s$, then, the mass of the fuel needed is
(Assume Avogadro’s number $ = 6 \times {10^{23}}mo{l^{ - 1}},{\text{ }}1eV = 1.6 \times {10^{ - 19}}J$)
A. 70.5 kg
B. 0.705 kg
C. 13.1 kg
D. 1.31 kg

Answer 1

Hint: In this question, first calculate total energy being released by making use of power and time given. Then we know the amount of energy that is released by one reaction, from which we can calculate the number reactions taking place per second. Then by making use of the atomic weight of uranium and the Avogadro’s number we can obtain the required answer.

Formula used:
The formula for power is given as
$P = \dfrac{W}{t}$

Complete step-by-step answer:
We are given that a ${U^{235}}$ nuclear reactor generates energy at a rate of $3.70 \times {10^7}J/s$. This means that the power of tis nuclear reactor is given as
$P = 3.70 \times {10^7}J/s$
We are given that the reactor operates for a duration given as
$t = 144 \times {10^4}s$
Now from the formula for power, we can calculate the total energy that is being generated by the reactor in this duration t. It can be calculated in the following way.
$W = P \times t = 3.70 \times {10^7} \times 144 \times {10^4} = 532.8 \times {10^{11}}J$
This is the total energy that the reactor is generating. We are given that for one nuclear reaction that amount of energy liberated has value equal to 185MeV. Therefore, we can calculate the number of fission reactions taking place in the reactor. This can be calculated in the following way.
$
  N = \dfrac{{{\text{Total energy generated}}}}{{{\text{Energy generated per reaction}}}} \\
   = \dfrac{{532.8 \times {{10}^{11}}}}{{185 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 1.8 \times {10^{24}}{\text{ reactions per sec}} \\
 $
Finally, the mass of uranium-235 required is equal to the product of the number of reactions with the atomic weight of uranium divided by the Avogadro’s number whose value we are given in the question. It can be calculated in the following way.
Mass $ = 235 \times \dfrac{{1.8 \times {{10}^{24}}}}{{6 \times {{10}^{23}}}} = 705g = 0.705kg$
Hence, the amount of U-235 required is 0.705kg and the correct answer is option B.

So, the correct answer is “Option B”.

Note: 1. By making the use of Avogadro’s number in the last step where by dividing the number of reactions with Avogadro’s number, we actually get the number of moles required uranium that are being consumed in the reaction.
2. It should also be noted that a very small amount of nuclear fuel is capable of producing a very large amount of energy.