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The United States uses $1.0 \times 10^{19} \mathrm{J}$ of electrical energy per year. If all this energy came from the fission of $^{235} \mathrm{U},$ which releases 200 MeV per fission event, (a) how many kilograms of $^{235}\mathrm{U}$ would be used per year; (b) how many kilograms of uraniumwould have to be mined per year to provide that much $^{235} \mathrm{U} ?$ (Recall that only 0.70$\%$ of naturally occurring uranium is $^{235} \mathrm{U} . )$

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a. $1.23 \times 10^{5} \mathrm{kg}$b. $1.76 \times 10^{7} \mathrm{kg}$

Physics 103

Chapter 30

Nuclear and High-Energy Physics

Atomic Physics

Nuclear Physics

Particle Physics

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

03:22

43.44. The United States u…

02:01

In 2010 , the United State…

02:30

The United States uses abo…

03:20

The energy consumed in one…

01:40

04:10

The world’s uranium supply…

01:51

M GO The energy consumed i…

08:54

The total energy consumed …

05:46

If each fission reaction o…

07:36

(a) If each fission reacti…

06:50

When 1.0 $\mathrm{kg}$ of …

05:27

04:16

When an atom of ${ }^{235}…

the United States produces one time scent of the 19 jewels of energy per year, which I call E. And the energy of the reactor is 200 MTV proficient event, which I call E prime, which is 3.2 times 10 to the minus 11 jewels. So using this information, we can go ahead and solve part, which is how many kilograms of uranium would be used for year in this situation. Okay, so first of all, let's calculate the, uh the number of moles required. So first, we're gonna find the number of nuclei in the event so this can come which is in this is just simply gonna be taking e and dividing it by you Prime. I take this ratio. I find that the number of nuclei is equal to, um, four point 375 times 10 to the floor. Excuse me? Times 10 to the 29. Sorry, then the number of moles that I have is just gonna be in divided by avocados number, which I I'm gonna call inside A. Of course, avocados Number is 6.2 times 10 to the 23rd. So taking this ratio giving the number of moles and this is equal to 7 to 6744 moles. Okay, now I can find a mass here because the mass we're gonna call him is equal to the ratio of end to end A which I just found, multiplied by the conversion ratio of uranium, which is 235 grams for mole. So the mass that I find here is equal to 1.7 times 10 to the eight grams or 1.7 times uh, 10 to the five kilograms mike and boxes and is my solution for part a Part B says that it wants me, uh, to assume that 70% of its naturally occurring uranium how much would have to be mine. So we're gonna let X b the number of kilograms of uranium required here. So X is gonna be multiplied by 0.7 great 0.7 divided by 100 because it's 1000.7% is equal to the mess that we just found. Therefore, solving for X, we find the X here would have to be equal to 2.4 times 10 to the seven kilograms we can box that in as our solution to be

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