Question

A 1000MW fission reactor consumes half of its fuel in 5 years. How much $_{92}^{235}U$ did it contain initially? Assuming that the reactor operates $80\,\% $ of the time, that all the energy generated arises from the fission of $_{92}^{235}U$ and that this nuclei is consumed only by the fission process.

Answer 1

Hint: In order to find the amount of uranium consumed first we need to find the total energy of the reactor and divide it by the energy released in the fission of $1\,kg$ of uranium.
The energy that is produced per fission of Uranium is $200MeV \cdot $. We can find the total energy by multiplying the energy per fission with the total number of nuclei in $1\,kg$ of uranium.
The initial amount will be twice the amount consumed.

Complete step by step answer:
It is given that the half-life of fuel is 5 years.
That is,
${t_{\dfrac{1}{2}}} = 5\,{\text{years}}$
The energy that is produced per fission of Uranium is $200MeV \cdot $
We need to find energy produced in $1\,kg$ uranium
For that let us first find the number of atoms in $1\,kg$ uranium. We can find it using the formula,
$n = \dfrac{{{N_A}}}{A} \times 1000$
Where ${N_A}$ is the Avogadro number whose value is ${N_A} = 6.022 \times {10^{23}}$, A is the mass number.
The total energy will be the product of energy released by a single nucleus and the total number of nuclei.
Therefore,
$E = \dfrac{{{N_A}}}{A} \times 1000 \times 200$
Now let us substitute the values,
$ \Rightarrow {E_{1\,kg}} = \dfrac{{6.022 \times {{10}^{23}}}}{{235}} \times 1000 \times 200\,Mev$
$ \Rightarrow {E_{1\,kg}} = \dfrac{{6.022 \times {{10}^{23}}}}{{235}} \times 1000 \times 200 \times 1.6 \times {10^{ - 13}}\,J$
$ \Rightarrow {E_{1\,kg}} = 8 \cdot 17 \times {10^{13}}J$
It is given that the reactor operates $80\% $ of time.
Therefore, time of operating,
$ \Rightarrow t = \dfrac{{80}}{{100}} \times 5$
$ \Rightarrow t = 4\,years$
Now let us calculate energy released in 5 years. Power of the reactor is given as $P = 1000\,MW = 1000 \times {10^6}\,W$
$ \Rightarrow t = 4 \times 365 \times 24 \times 60 \times 60\,s$
We know that energy is power multiplied by time.
${E_R} = P \times t$
On substituting the given values we get,
$ \Rightarrow {E_R} = 1000 \times {10^6} \times 60 \times 60 \times 24 \times 365 \times 4\,J$
We need to find the amount of uranium consumed. This can be found out by dividing total energy by the energy released by $1\,kg$ uranium. That is,
 ${\text{Amount}}\,{\text{consumed}} = \dfrac{{{E_R}}}{{{E_{1kg}}}}$
On substituting the values we get,
$ \Rightarrow {\text{Amount}}\,{\text{consumed}} = \dfrac{{1000 \times {{10}^6} \times 60 \times 60 \times 24 \times 365 \times 4}}{{8 \cdot 17 \times {{10}^{13}}}}$

$\therefore {\text{Amount}}\,{\text{consumed = }}1544\,kg$

So, the amount consumed in kg is $1544\,kg$

The initial amount will be two times 1544. Because it is given that half of the Uranium is consumed. Which means initially the amount will be double the amount consumed. That is, $2 \times 1544 = 3088\,kg$.

Note:
Remember that the value $200\,MeV$ is the energy of fission of a single uranium nucleus. In order to find the energy released by $1\,kg$ find the number of nucleus in $1\,kg$ using the equation
$n = \dfrac{{{N_A}}}{A} \times 1000$. so total energy for $1\,kg$ will be energy per fission multiplied by total number of nucleus in $1\,kg$.