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Chapter 37: Problem 86

(a) How much energy is released in the explosion of a fission bomb containing \(3.0 \mathrm{~kg}\) of fissionable material? Assume that \(0.10 \%\) of the mass is converted to released energy. (b) What mass of TNT would have to explode to provide the same energy release? Assume that each mole of TNT liberates 3.4 MJ of energy on exploding. The molecular mass of TNT is \(0.227 \mathrm{~kg} / \mathrm{mol}\). (c) For the same mass of explosive, what is the ratio of the energy released in a nuclear explosion to that released in a TNT explosion?

Short Answer

Expert verified
The fission bomb releases 2.7 脳 10鹿鈦� J, needing 1.80 脳 10鈦� kg of TNT for the same energy. The energy ratio is 6.0 脳 10鈦�.

Step by step solution

01

Calculate Energy Released by the Fission Bomb

First, we need to determine the amount of mass converted into energy. Given 3.0 kg of fissionable material, and assuming 0.10% is converted to energy, we calculate this as follows:\[ \text{Mass converted} = 3.0 \, \text{kg} \times 0.001 = 0.003 \, \text{kg} \]Now, using Einstein's mass-energy equivalence principle \( E = mc^2 \), where \( c = 3.00 \times 10^8 \, \text{m/s} \) is the speed of light, the energy released is:\[ E = 0.003 \, \text{kg} \times (3.00 \times 10^8 \, \text{m/s})^2 = 2.7 \times 10^{14} \, \text{J} \]
02

Calculate Mass of TNT Required for Equivalent Energy

To find the equivalent mass of TNT, knowing that 1 mole of TNT liberates 3.4 MJ (or 3.4 \( \times 10^6 \, \text{J} \)) and the molecular mass of TNT is 0.227 kg/mol, we first express the energy released by the bomb in terms of TNT energy:\[ \text{Energy per mol of TNT} = 3.4 \times 10^6 \, \text{J} \]The number of moles of TNT required is:\[ \text{Moles of TNT} = \frac{2.7 \times 10^{14} \, \text{J}}{3.4 \times 10^6 \, \text{J/mol}} = 7.94 \times 10^7 \, \text{mol} \]The mass of TNT is then calculated as:\[ \text{Mass of TNT} = 7.94 \times 10^7 \, \text{mol} \times 0.227 \, \text{kg/mol} = 1.80 \times 10^7 \, \text{kg} \]
03

Calculate Energy Ratio of Nuclear to TNT Explosion

To find the ratio of energy release, we compare the energy per unit mass for each explosive. Using 3.0 kg for the nuclear material:\[ E_{\text{nuclear}} = \frac{2.7 \times 10^{14} \, \text{J}}{3.0 \, \text{kg}} = 9.0 \times 10^{13} \, \text{J/kg} \]For TNT, knowing each mole is 0.227 kg:\[ E_{\text{TNT}} = \frac{3.4 \times 10^6 \, \text{J}}{0.227 \, \text{kg}} = 1.50 \times 10^7 \, \text{J/kg} \]Finally, the energy ratio is:\[ \text{Ratio} = \frac{9.0 \times 10^{13} \, \text{J/kg}}{1.50 \times 10^7 \, \text{J/kg}} = 6.0 \times 10^6 \]
04

Conclusion

The energy ratio indicates that the nuclear explosion releases 6 million times more energy per kilogram compared to TNT.

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Key Concepts

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

Mass-Energy Equivalence
One of the most groundbreaking ideas in physics is the principle of mass-energy equivalence, introduced by Albert Einstein. This idea stems from his famous equation \( E = mc^2 \), which explains the relationship between mass (\( m \)) and energy (\( E \)), with \( c \) being the speed of light in a vacuum. In essence, this equation tells us that mass can be converted into energy and vice versa.

In the context of nuclear fission, a small amount of mass is transformed into a large amount of energy. If you take 3.0 kg of fissionable material and convert 0.10% of it into energy, as illustrated in the exercise, a mere 0.003 kg converts into an incredible \( 2.7 \times 10^{14} \) joules of energy. This transformation occurs because the speed of light is so vast (\( 3.00 \times 10^8 \) m/s), amplifying the energy that results from even minute mass changes.

This concept of transforming mass into energy is central to how nuclear bombs release their explosive energy, making them incredibly powerful compared to traditional explosives like TNT.
Nuclear Fission
Nuclear fission is a process where the nucleus of an atom splits into two or more smaller nuclei, along with the release of a significant amount of energy. This splitting occurs because certain heavy nuclei, like uranium or plutonium, become unstable when they absorb a neutron.

When these nuclei split, they release additional neutrons and energy, which can then strike other heavy nuclei, causing them to split as well. This chain reaction is what powers nuclear fission bombs.
  • Fission releases a colossal amount of energy due to the conversion of mass into energy, as per the mass-energy equivalence principle.
  • This released energy is astoundingly greater compared to the energy from chemical reactions as in conventional explosives.
In the exercise example, nuclear fission is responsible for turning a small portion of the nuclear material into immense energy output, calculated using Einstein's equation. This highlights how nuclear fission is a powerful energy process due to the efficiency of mass-energy transformation.
Explosive Energy Comparison
The comparison between the explosive energy released by nuclear fission and that by traditional explosives like TNT is striking. In the exercise, we see just how much more energy nuclear processes can unleash compared to chemical reactions.

Let's break this down:
  • For the fission bomb, the calculation showed \( 2.7 \times 10^{14} \) joules of energy released from only 3 kg of material.
  • On the other hand, to match this energy release with TNT, an enormous 18 million kg of TNT would be required due to its much lower energy per unit mass energy (1.50 \times 10^7 J/kg).
This disparity in energy release per kilogram is evident by the energy ratio established in the exercise: nuclear explosions yield approximately 6 million times more energy per kilogram than TNT explosions.

Understanding this comparison aids in grasping the sheer power of nuclear energy and its implications for both energy generation and weaponry.

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Most popular questions from this chapter

A particle with mass \(m\) has speed \(c / 2\) relative to inertial frame S. The particle collides with an identical particle at rest relative to frame \(S\). Relative to \(S\), what is the speed of a frame \(S^{\prime}\) in which the total momentum of these particles is zero? This frame is called the center of momentum frame.

To eight significant figures, what is speed parameter \(\beta\) if the Lorentz factor \(\gamma\) is (a) \(1.0100000\), (b) \(10.000000\), (c) \(100.00000\), and (d) \(1000.0000 ?\)

The total energy of a proton passing through a laboratory apparatus is \(10.611 \mathrm{~nJ} .\) What is its speed parameter \(\beta ?\) Use the proton mass given in Appendix B under "Best Value," not the commonly remembered rounded number.

An alpha particle with kinetic energy \(7.70 \mathrm{MeV}\) collides with an \({ }^{14} \mathrm{~N}\) nucleus at rest, and the two transform into an \({ }^{17} \mathrm{O}\) nucleus and a proton. The proton is emitted at \(90^{\circ}\) to the direction of the incident alpha particle and has a kinetic energy of \(4.44 \mathrm{MeV}\). The masses of the various particles are alpha particle, \(4.00260 \mathrm{u} ;{ }^{14} \mathrm{~N}\). \(14.00307\) u; proton, \(1.007825 \mathrm{u}\); and \({ }^{17} \mathrm{O}, 16.99914 \mathrm{u}\). In \(\mathrm{MeV}\), what are (a) the kinetic energy of the oxygen nucleus and (b) the \(Q\) of the reaction? (Hint: The speeds of the particles are much less than \(c .\) )

A certain particle of mass \(m\) has momentum of magnitude \(m c\). What are (a) \(\beta\), (b) \(\gamma\), and (c) the ratio \(K / E_{0}\) ?
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