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Chapter 19: Problem 67

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
Approximately 940 grams of ${ }^{235} \mathrm{U}$ undergoes fission in the atomic bomb.

Step by step solution

01

Calculate the energy released by the atomic bomb

To calculate the energy released by the atomic bomb, we need to multiply the energy released by the detonation of 1 ton of TNT (4 × 10^9 J) by the total number of equivalent tons of TNT (20,000 tons). This can be written as: Energy released = (20,000 tons)(4 × 10^9 J/ton)
02

Convert the energy released to moles of uranium-235

Now that we know the energy released by the atomic bomb, we want to find out how many moles of uranium-235 would release this energy when undergoing fission. We do this by dividing the total energy released by the bomb by the energy released per mole of uranium-235 fission (2 × 10^13 J/mol). Let this value be N: N (moles of uranium-235) = (Energy released) / (2 × 10^13 J/mol)
03

Calculate the mass of uranium-235

Now that we know the number of moles of uranium-235 that undergoes fission, we need to find the mass of this amount of uranium. To do that, we multiply the number of moles (N) by the molar mass of uranium-235 (235 g/mol). This can be written as: Mass of uranium-235 = N × (235 g/mol) Combine all steps to find the solution: Energy released = (20,000 tons)(4 × 10^9 J/ton) = 8 × 10^13 J N = (8 × 10^13 J) / (2 × 10^13 J/mol) = 4 moles of uranium-235 Mass of uranium-235 = 4 × 235 g/mol = 940 g Hence, approximately 940 grams of 235U undergoes fission in this atomic bomb.

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Most popular questions from this chapter

Much of the research on controlled fusion focuses on the problem of how to contain the reacting material. Magnetic fields appear to be the most promising mode of containment. Why is containment such a problem? Why must one resort to magnetic fields for containment?

Iodine-131 has a half-life of \(8.0\) days. How many days will it take for \(174 \mathrm{~g}\) of \({ }^{131}\) I to decay to \(83 \mathrm{~g}\) of \({ }^{131} \mathrm{I}\) ?

A reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is \(15.0\) seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would

The mass ratios of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9}\) years \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e}(89.3 \%)\) a. Why are \({ }^{40} \mathrm{Ar} /{ }^{40} \mathrm{~K}\) ratios used to date materials rather than \({ }^{40} \mathrm{Ca} /{ }^{40} \mathrm{~K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an \({ }^{40} \mathrm{Ar} /{ }^{40} \mathrm{~K}\) ratio of \(0.95\). Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \({ }^{40} \mathrm{Ar}\) escaped from the sample?

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 \(1 \mathrm{~g}\) radium in \(1 \mathrm{~s}\) ). a. What mass of \(\mathrm{Na}_{2}{ }^{38} \mathrm{SO}_{4}\) has an activity of \(10.0 \mathrm{mCi}\) ? Sulfur- 38 has an atomic mass of \(38.0 \mathrm{u}\) and a half-life of \(2.87 \mathrm{~h}\). b. How long does it take for \(99.99 \%\) of a sample of sulfur- 38 to decay?
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