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

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

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Anonym0us (Student, UG1 Grade)

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I found an answer from courses.lumenlearning.com

Fission | Physics

Fission is the opposite of fusion and releases energy only when heavy nuclei are split. ... The amount of energy per fission reaction can be large, even by nuclear standards. ... given the atomic masses to be m(²³⁸U) = 238.050784 u, m(⁹⁵Sr) ... fissions produced by neutrons by having a large amount of fissionable material.

For more information, see Fission | Physics

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Krishna (last edited 6 months ago) (Expert, Educator @Qalaxia)

Given that

Fission reactor power P = 1000 MW = 10^3* 10^{6} W = 10^{9} W

Half life of the ^{235}_{92} U, T_{\frac{1}{2}} = 5 years

Suppose 80% of the time the reactor operates = 80% of T_{\frac{1}{2}} = \frac{80}{100} * 5 = 4 years

\text{ 4 years } = 4*360*24 *60*60 . seconds

= 124416000 seconds

Step 1: Finding the number of atoms in the 1 kg fission matter

Avogadro number NA = 6.023 * 10^{23}

Atomic mass of ^{235}_{92} U =. 235

Mass of a Uranium-235 m = 1kg = 10^{3} g

Number of atoms in the 1 kg _{92}^{235}U,\ \ N=\frac{NA}{A}*m

N = \frac{6.023 * 10^{23}}{235} * 10^{3}

Step 2: Determine the total energy released by 1 kg mass of substance

The average amount of energy released per nuclear fusion = 200 MeV

E = \frac{6.023 * 10^{23}}{235} * 10^{3}* 200 MeV

E = 5.106 * 10^{26} MEV

E = 5.106 * 10^{26} * 1.6 * 10^{-13} Joules

E = 8.1696 * 10^{13} Joules

Step 3: Calculating the total energy consumed by substance in half life by fission reactor.

\text{ Energy consumed }= \frac{\text{ Energy of reactor released in 5 years }}{\text{ Energy release in 1 kg of substance }}

= \frac{124416000 * 10^{9}}{ 8.1696 * 10^{13} }

= 1522.9 kg

Hence, energy consumed by Uranium-235 = 1522.9 kg

Since the half life of Uranium-235 consumed = 1522.9 kg, the initially the amount will be doubled = 2 * 1522.9 = 3045.8 kg