A 1000 MW fission reactor consumes half of its fuel in 5.00 y. 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.

Solution

Given,

Power of reactor = 1000 MW = 10³MW
= 10⁹W
= 10⁹ Js^-1Energy generated by reactor in 5 Years = 5 x 365 x 24 x 60 x 60 x 10⁹J

Average energy generated = 200 MeV
= 200 x 1.6 x 10^-13J

Number of fission taking place or number of U²³⁵nuclei required,
$= \frac{5 \times 365 \times 24 \times 60 \times 60 \times 10^{9}}{200 \times 1.6 \times 10^{- 13}}$

$= 8.2125 \times 10^{26} \times 6 = 49.275 \times 10^{26}$

Mass of 6.023 x 10²³nuclei of U = 235 gm = 235 x 10^-3 kg

Mass of 8.2125 x 10²⁶ nuclei of U,
$= \frac{235 \times 10^{- 3}}{6.023 \times 10^{23}} \times 6 \times 8.2125 \times 10^{26} = 1932 kg$
$\frac{1}{2} of fuel = 1932 kg Total fuel = 3864 kg .$

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Wave Optics

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. 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.

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