Question

Nuclear reactor in which ${\text{U}} - 235$ is used as fuel. Uses $2kg$ of ${\text{U}} - 235$ in $30$ days. Then, power output of the reactor will be (given energy released per fission $= 185{\text{MeV}}$ ).
(A) $43.5{\text{MW}}$
(B) $58.5{\text{MW}}$
(C) $73.1{\text{MW}}$
(D) $69.6{\text{MW}}$

Answer 1

To solve this question, we have to calculate the number of moles of the given mass of ${\text{U}} - 235$ , from which the number of atoms can be calculated. The number of fissions will be equal to the number of atoms of ${\text{U}} - 235$ , so the energy released from the given mass of ${\text{U}} - 235$ can be calculated from the given value of energy per fission. Finally, from the given time, the power can be calculated.

Formula used: The formulae used for solving this question are given by
$n = \dfrac{M}{{{M_0}}}$ , here $n$ is the number of moles, $M$ is the mass, and ${M_0}$ is the molar mass.
$P = \dfrac{E}{t}$ , here $P$ is the power, $E$ is the energy, and $t$ is the time.

Complete step-by-solution
According to the question, the energy released from the fission of one atom of ${\text{U}} - 235$ is given.
So for obtaining the energy released from the fission of the given mass of ${\text{U}} - 235$ , we first have to calculate the number of atoms of ${\text{U}} - 235$ .
We know that the number of moles is given by
$n = \dfrac{M}{{{M_0}}}$
Now, according to the question, the given mass of ${\text{U}} - 235$ is equal to $2kg$ . Also, the designation ${\text{U}} - 235$ means that the atomic mass of ${\text{U}} - 235$ is equal to $235g/mol$ . Therefore we substitute $M = 2kg = 2000g$ and ${M_0} = 235g/mol$ in the above equation to get
$n = \dfrac{{2000}}{{235}}$
Now, we know that the number of atoms is equal to Avogadro number times the number of moles. So the number of atoms of ${\text{U}} - 235$ is given by
$N = 6.022 \times {10^{23}} \times \dfrac{{2000}}{{235}}$
On solving we get
$N = 5.12 \times {10^{24}}$
According to the question, the energy released per fission of ${\text{U}} - 235$ is equal to $185{\text{MeV}}$ . Therefore, the energy released from the fission of $5.12 \times {10^4}$ is given by
$E = 5.12 \times {10^{24}} \times 185{\text{MeV}}$
On solving we get
$E = 947.2{\kern 1pt} \times {10^{24}}{\text{MeV}}$
We know that
$1{\text{eV}} = 1.6 \times {10^{ - 19}}{\text{J}}$
Therefore, we get
$E = 947.2 \times {10^{24}} \times 1.6 \times {10^{ - 19}}{\text{MJ}}$
$\Rightarrow E = 1515.52 \times {10^5}{\text{MJ}}$ …………..(1)
Now, the power is given by
$P = \dfrac{E}{t}$ …………..(2)
According to the question, the fission takes place in $30$ days. So the time is
$t = 30days$
We know that there are $24$ hours in one day. So we have
$t = 30 \times 24hrs$
Now, one hour equals sixty minute, and one minute equals sixty seconds. So the time is seconds is given by
$t = 30 \times 24 \times 60 \times 60s$ …………..(3)
Putting (1) and (3) in (2) we have
$P = \dfrac{{1515.52 \times {{10}^5}{\text{MJ}}}}{{30 \times 24 \times 60 \times 60s}}$
On solving we get
$P = 58.5{\text{MW}}$
Thus, the power output of the nuclear reactor is equal to $58.5{\text{MW}}$ .
Hence, the correct answer is option B.

Note
The nuclear reactors are widely used for the purpose of electricity generation. They are used in preparing isotopes which are used in medicines and industries. Also, they are very much popular for their use in the production of nuclear weapons.