Question

In a nuclear reactor, ${}^{235}U$ undergoes fission liberating 200 MeV of energy. The reactor has 10% efficiency and produces 1000MW power. If the reactor is to function for 10 years then find the total mass of Uranium required.
(A) $36.5\times {10}^{3}kg$
(B) $36\times {10}^{2}kg$
(C) $39.5\times {10}^{3}kg$
(D) $38.2\times {10}^{3}kg$

Answer 1

Hint: Uranium undergoes fission to generate energy in a nuclear reactor. The mass required can be back-calculated from the energy is produced. Sample nuclear reaction used in nuclear reactor:
$\text{Nuclear fission reaction: }{}_0^{1}n+{}_{92}^{235}U\to {}_{56}^{141}Ba+{}_{36}^{92}U+3{}_0^{1}n+Energy$ (it is just sample of uranium decay reaction in nuclear reactor. It is not an exact reaction for this question as it just says energy released and efficiency not the product of reaction).

Formula used:
Energy:
$E=Pt$ ……(1)
where,
$P$ is the power in Watts
$t$ is the time.
Input energy:
$E_{in}={\displaystyle \frac{E_{out}}{Efficiency}}$ ……(2)
where,
$E_{out}$ is the output energy
Conversion of eV to J:
$1eV=1.6\times {10}^{-19}J$ ……(3)
Convert number of fissions to number of moles:
$n={\displaystyle \frac{no.}{N_A}}$ ……(4)
where,
$no.$= number of fissions
$N_A$ is the Avogadro number
Mass:
$m=n\times M$ ……(5)
where,
$n$ is the number of moles,
$M$ is mass per mole (molar mass).

Complete step by step solution:
Given:
1. Energy liberated by 1 fission of ${}^{235}U$($E_{fission}$) = $200MeV$
2. efficiency of reactor = 10%
3. Power produced = 1000MW
4. Number of years reactor functions = 10

To find: The expression for energy stored in the capacitor.
We know,
Nuclear fission reaction:
${}_{92}^{235}U+{}_0^{1}n\to {}_{56}^{141}Ba+{}_{36}^{92}Kr+3{}_0^{1}n+Heat$ (just for information, this question doesn’t require reaction use)

Step 1 of 3:
Calculate the energy which the reactor would supply for 10 years using eq (1):
$$\begin{gathered} E=1000\times {10}^6\times (10\times 365\times 24\times 3600s) \\ E=3.15\times {10}^{17}J \end{gathered}$$
This is the output energy.
As the efficiency is only 10%, the actual energy which had to be supplied would be calculated by eq (2):
$$\begin{gathered} E_{in}={\displaystyle \frac{3.15\times {10}^{17}}{{\displaystyle \frac{10}{100}}}} \\ {E}_{in}=3.15\times {10}^{18}J \end{gathered}$$

Step 2 of 3:
Convert the energy produced by fission of 1 ${}^{235}U$ from $MeV$ to $J$ using eq (3):
$$\begin{gathered} E_{fission}=E_{in}\times 1.6\times {10}^{-19} \\ {E}_{fission}=200\times {10}^6\times 1.6\times {10}^{-19} \\ {E}_{fission}=3.2\times {10}^{-11}J \end{gathered}$$

Step 3 of 3:
Find the total number of fissions required by:
$$\begin{gathered} no.={\displaystyle \frac{E_{in}}{E_{fission}}} \\ no.={\displaystyle \frac{3.15\times {10}^{18}}{3.2\times {10}^{-11}}} \\ no.\approx 9.8\times {10}^{28} \end{gathered}$$
Convert this number of fissions to number of moles using eq (4):
$$\begin{gathered} n={\displaystyle \frac{9.8\times {10}^{28}}{6.02\times {10}^{26}}} \\ n=163.7 \end{gathered}$$
Molar mass of ${}^{235}U$ is 235 kg/mol. Mass required for these many moles is given by eq (5):
$$\begin{gathered} m=163.7\times 235 \\ m=38.2\times {10}^{3}kg \end{gathered}$$

$\therefore$ The required mass of Uranium is $38.2\times {10}^{3}kg$. Hence, option (D) is correct.

Note:
Both nuclear fission and fusion release huge amounts of energy which is further converted into Electrical Energy through a controlled chain reaction in nuclear reactors.
It is good to read about the Manhattan Nuclear project. This project initially focused on producing the first Nuclear weapons during world war II. But later, the idea of nuclear reaction was used to transform its immense energy into generating electricity.