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A small atomic bomb releases energy equivalent to the detonation of 20,000 tons of TNT; a ton of TNT releases \(4 \times 10^{9} \mathrm{J}\) of energy when exploded. Using \(2 \times 10^{13} \mathrm{J} / \mathrm{mol}\) as the energy released by fission of \(^{235} \mathrm{U},\) approximately what mass of \(^{235} \mathrm{U}\) undergoes fission in this atomic bomb?

Short Answer

Expert verified
To find the mass of U-235 undergoing fission in a small atomic bomb, we first calculate the total energy released, which is \(4 \times 10^9 J \times 20,000\). Then, find the number of moles of U-235 required by dividing the energy released by the atomic bomb by the energy released per mole of U-235, which is \(\frac{4\times10^9 J \times 20,000}{2\times10^{13} J/mol}\). Finally, convert the number of moles to mass using the molar mass of U-235 (238 g/mol), resulting in \(\frac{4\times10^9 J \times 20,000}{2\times10^{13} J/mol}\) x 238 g/mol.

Step by step solution

01

Calculate the total energy released by the atomic bomb

To calculate the total energy released by the atomic bomb, we first need to multiply the energy released by a single ton of TNT by the 20,000 tons of TNT, which is equivalent to the atomic bomb's energy: Energy released by atomic bomb = Energy released by 1 ton of TNT x 20,000 tons of TNT = \(4 \times 10^9 J \times 20,000\)
02

Compute the number of moles of U-235 required

Now, we know the energy released by the atomic bomb, and we know the energy released per mole of U-235 during fission. So, we can compute the number of moles of U-235 required using the following equation: Number of moles of U-235 = Energy released by atomic bomb / Energy released per mole of U-235 Number of moles of U-235 = \(\frac{4\times10^9 J \times 20,000}{2\times10^{13} J/mol}\)
03

Calculate the mass of U-235

Finally, we need to convert the number of moles of U-235 to the mass of U-235. To do this, we will multiply the number of moles by the molar mass of U-235 (which is 238 g/mol): Mass of U-235 = Number of moles of U-235 x Molar mass of U-235 Mass of U-235 = \(\frac{4\times10^9 J \times 20,000}{2\times10^{13} J/mol}\) x 238 g/mol

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Key Concepts

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

Energy Calculation
Understanding energy calculation in nuclear fission involves determining the total energy released by an atomic bomb. In this case, the energy is compared to the explosion energy of 20,000 tons of TNT. Each ton of TNT produces \(4 \times 10^{9}\) joules of energy. Therefore, multiplying this by 20,000 gives the total energy released by the bomb. This calculation is essential for understanding how much energy is generated and how it translates into a physical equivalent. Recognizing energy conversion is key as it reveals the potent power of nuclear fission compared to traditional explosives.
Uranium-235
Uranium-235 (U-235) is a crucial isotope in nuclear fission processes. It has unique properties making it highly suitable for sustaining a nuclear chain reaction. During fission, U-235 nuclei absorb neutrons, become unstable, and split into smaller nuclei while releasing enormous amounts of energy. This energy release is specified as \(2 \times 10^{13} \text{J/mol}\) for U-235, illustrating its high energy potential per mole. Understanding these properties helps us appreciate why U-235 is favored in nuclear reactors and weapons. Its availability and characteristics make it a valuable but carefully managed nuclear resource.
Molar Mass Calculation
Calculating the molar mass is crucial in converting from moles to grams, providing a tangible way to measure substances. For Uranium-235, the molar mass is 238 g/mol. This specific calculation enables us to determine the actual mass needed for fission once we have calculated the number of moles. To find the mass of U-235 required, multiply the number of moles by 238 g/mol. These computations bridge the gap between theoretical and practical applications of nuclear materials, making it easier to visualize and measure amounts in real-world scenarios.

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Most popular questions from this chapter

In each of the following radioactive decay processes, supply the missing particle. a. \(^{60} \mathrm{Co} \rightarrow^{60} \mathrm{Ni}+?\) b. \(^{97} \mathrm{Tc}+? \rightarrow^{97} \mathrm{Mo}\) c. \(^{99} \mathrm{Tc} \rightarrow^{99} \mathrm{Ru}+?\) d. \(^{239} \mathrm{Pu} \rightarrow^{235} \mathrm{U}+?\)

The rate constant for a certain radioactive nuclide is \(1.0 \times\) \(10^{-3} \mathrm{h}^{-1} .\) What is the half-life of this nuclide?

The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from 1 g radium in 1 s). A 1.7 -mL sample of water containing tritium was injected into a 150 -lb person. The total activity of radiation injected was \(86.5 \mathrm{mCi}\). After some time to allow the tritium activity to equally distribute throughout the body, a sample of blood plasma containing \(2.0 \mathrm{mL}\) water at an activity of \(3.6 \mu \mathrm{Ci}\) was removed. From these data, calculate the mass percent of water in this 150 -lb person.

At a flea market, you've found a very interesting painting done in the style of Rembrandt's "dark period" (1642-1672). You suspect that you really do not have a genuine Rembrandt, but you take it to the local university for testing. Living wood shows a carbon- 14 activity of 15.3 counts per minute per gram. Your painting showed a carbon- 14 activity of 15.1 counts per minute per gram. Could it be a genuine Rembrandt?

A \(0.10-\mathrm{cm}^{3}\) sample of a solution containing a radioactive nuclide \(\left(5.0 \times 10^{3}\) counts per minute per milliliter) is injected \right. into a rat. Several minutes later \(1.0 \mathrm{cm}^{3}\) blood is removed. The blood shows 48 counts per minute of radioactivity. Calculate the volume of blood in the rat. What assumptions must be made in performing this calculation?

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