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The United States uses about 1.4 \(\times\) 10\(^1$$^9\) J of electrical energy per year. If all this energy came from the fission of \(^2$$^3$$^5\)U, which releases 200 MeV per fission event, (a) how many kilograms of \(^2$$^3$$^5\)U would be used per year, and (b) how many kilograms of uranium would have to be mined per year to provide that much \(^2$$^3$$^5\)U? (Recall that only 0.70% of naturally occurring uranium is \(^2$$^3$$^5\)U.)

Short Answer

Expert verified
For part (a), 170.57 kg of \\(^{235}U\\) is used per year. For part (b), approximately 24367.1 kg of uranium would need to be mined each year.

Step by step solution

01

Convert Energy Released per Fission to Joules

First, we need to convert the energy released per fission from MeV to joules. Given that 200 MeV is released per fission, we use the conversion factor 1 MeV = 1.60218 脳 10鈦宦孤 J.\[200 \text{ MeV} \times 1.60218 \times 10^{-13} \text{ J/MeV} = 3.20436 \times 10^{-11} \text{ J}\]
02

Determine Total Fissions Needed

To find out how many fission events are needed to produce the annual electricity needs, divide the total energy requirement by the energy per fission event:\[\frac{1.4 \times 10^{19} \text{ J}}{3.20436 \times 10^{-11} \text{ J/fission}} \approx 4.37 \times 10^{29} \text{ fissions}\]
03

Calculate Mass of Uranium Used

Calculate the mass of \(^{235}\text{U}\) atoms needed given each fission uses one atom. The molar mass of \(^{235}\text{U}\) is approximately 235 g/mol. Use Avogadro's number for the conversion:\[\frac{4.37 \times 10^{29} \text{ fissions}}{6.022 \times 10^{23} \text{ atoms/mol}} \times 235 \text{ g/mol} \approx 1.7057 \times 10^5 \text{ g}\]Convert grams to kilograms:\[1.7057 \times 10^2 \text{ kg}\]
04

Calculate Total Uranium Mined

Since only 0.7% of natural uranium is \(^{235}\text{U}\), calculate the total mass of uranium mined. Use the percentage to find the total initial uranium:\[\frac{1.7057 \times 10^2 \text{ kg}}{0.007} \approx 24367.1 \text{ kg}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

uranium-235
Uranium-235 is a radioactive isotope of uranium that is commonly used as fuel in nuclear reactors and weapons. It has 92 protons and 143 neutrons in its nucleus, making the total mass number 235. This isotope is special because it is one of the few materials that can undergo nuclear fission, the process by which an atomic nucleus splits into smaller parts, releasing a significant amount of energy.
Uranium-235 is not very abundant in nature. Only about 0.7% of uranium found in minerals is the 235 isotope, the rest being mostly uranium-238. This rarity makes it necessary to enrich natural uranium to increase the concentration of uranium-235 in nuclear fuel applications.
When uranium-235 absorbs a neutron, it becomes unstable and splits into two smaller nuclei along with a few additional neutrons and a large amount of energy in the form of kinetic energy. This reaction is what powers nuclear reactors and atomic bombs.
energy conversion
Energy conversion in nuclear fission involves transforming the energy stored in the atomic nucleus into usable forms like heat and electricity. Each time a uranium-235 nucleus undergoes fission, it releases around 200 MeV (million electron volts) of energy.
To make this energy comprehensible for practical uses, it's converted into joules 鈥 the standard energy unit. For instance, in our specific case, the 200 MeV released per fission event equates to approximately \(3.20436 \times 10^{-11}\) Joules.
When dealing with large amounts of energy like those required by a country each year, extensive quantities of fission events are necessary. As calculated, providing 1.4 脳 10\(^{19}\) J of energy annually would require about 4.37 脳 10\(^{29}\) fission reactions. This energy can then be used to produce electricity for millions of homes.
Avogadro's number
Avogadro's number is a fundamental constant in chemistry and physics, often used to calculate the number of atoms or molecules in a given amount of substance. It is defined as 6.022 脳 10\(^{23}\) entities per mole. This number allows us to connect the microscopic world of atoms and molecules to the macroscopic world.
In nuclear fission calculations, Avogadro's number helps determine the mass of uranium-235 needed for a specific number of fission events. For instance, knowing how many fission reactions are required, we can use this constant to calculate how much uranium-235 is needed, ensuring a seamless link between atomic-scale reactions and large-scale energy production.
This is vital for understanding how much material we need on a practical level, such as in the mining and enrichment of uranium for nuclear fuel production.
molar mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For uranium-235, its molar mass is approximately 235 g/mol.
This concept is crucial in nuclear calculations as it allows us to convert between mole-based quantities and mass-based quantities, which are more practical for large-scale operations. By knowing the molar mass of uranium-235, we can determine how many grams or kilograms are needed to sustain a given number of fission reactions.
In solving the exercise, calculating the total mass of uranium-235 utilized required knowing that one mole contains approximately 235 grams. Using the number of fissions needed annually, we converted this to gram mass, finally resulting in the calculation of uranium required per year. Molar mass thus bridges the gap between chemical quantities and real-world mass units.

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Most popular questions from this chapter

(a) If a chest x ray delivers 0.25 mSv to 5.0 kg of tissue, how many \(total\) joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 mSv per year at sea level. Assuming an RBE of 1, how many rem and rads is this dose, and how many joules of energy does a 75-kg person receive in a year? (c) How many chest x rays like the one in part (a) would it take to deliver the same \(total\) amount of energy to a 75-kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

The radioactive nuclide \(^1$$^9$$^9\)Pt has a half-life of 30.8 minutes. A sample is prepared that has an initial activity of 7.56 \(\times\) 10\(^1$$^1\) Bq. (a) How many \(^1$$^9$$^9\)Pt nuclei are initially present in the sample? (b) How many are present after 30.8 minutes? What is the activity at this time? (c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared.

\(\textbf{To Scan or Not to Scan?}\) It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans) using x rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 mSv, applied to the \(whole\) \(body\). By contrast, a chest x ray typically administers 0.20 mSv to only 5.0 kg of tissue. How many chest x rays would deliver the same total amount of energy to the body of a 75-kg person as one whole-body scan?

As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 400 \(\mu\)Ci of \(^1$$^3$$^1\)Ba, which has a half-life of 12 days. (a) What mass of \(^1$$^3$$^1\)Ba was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00 \(\mu\)Ci. How long will the lab have to be closed?

What particle (a particle, electron, or positron) is emitted in the following radioactive decays? (a) \(^{27}_{14}Si\) \(\rightarrow\) \(^{27}_{13}Al\); (b) \(^{238}_{92}U\) \(\rightarrow\) \(^{234}_{90}Th\); (c) \(^{74}_{33}As\) \(\rightarrow\) \(^{74}_{34}Se\).

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