Question

A $1000MW$ fission reactor consumes half of its fuel in $5.00y$. How much ${}_{92}^{235}U$ did it contain initially? Assume that the reactor operates $80\%$ of the time, that all the energy generated arises from the fission of ${}_{92}^{235}U$ and that this nuclide is consumed only by the fission process.

Answer 1

The half life of the fuel is $5years$

we know that the energy produced in fission of Uranium is $200MeV$.

So, energy produced by 1kg of Uranium:

$E=\frac{6.022\times {10}^{23}}{235}\times 1000\times 200$

$\approx 8.17\times {10}^{13}J$

Since the reactor operate $80$ of time, so time of operation is $4years$.

The energy produced is:

$E_r=1000\times 1066\times 4\times 365\times 24\times 60\times 60$

The mass required for producing energy is:

$m=\frac{8.17\times {10}^{13}J}{1000\times 1066\times 4\times 365\times 24\times 60\times 60}$

$=1544kg$

Since this is the amount consumed and initial amount should have been double of this. So, the initial amount of Uranium would have been $3088kg$