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Chapter 14: Problem 42

The hydrogen bomb represents an effort to create a similar process to what takes place in the core of the Sun. The energy released by a 5-megaton hydrogen bomb is \(2 \times 10^{16} \mathrm{J}\) a. This textbook, \(21^{\text {th }}\) Century Astronomy, has a mass of about 1.6 kg. If all of its mass were converted into energy, how many 5 -megaton bombs would it take to equal that energy? b. How much mass did Earth lose each time a 5-megaton hydrogen bomb was exploded?

Short Answer

Expert verified
a. 7.2 bombsb. 0.22 kg per bomb

Step by step solution

01

Understand the Problem

We need to determine how many 5-megaton hydrogen bombs would be equivalent to the energy produced by converting the mass of a textbook into energy (part a) and calculate the mass loss of Earth per hydrogen bomb explosion (part b).
02

- Calculate Energy from Mass

Use Einstein's mass-energy equivalence formula: \[ E = mc^2 \]Here, the mass \( m = 1.6 \text{ kg} \)and the speed of light \( c = 3 \times 10^8 \text{ m/s} \). Plugging the values in:\[ E = 1.6 \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 = 1.6 \times 9 \times 10^{16} \text{ J} = 1.44 \times 10^{17} \text{ J} \]
03

- Calculate Number of Bombs Equivalent to Book's Energy

The energy of one 5-megaton bomb is given as \(2 \times 10^{16} \text{ J} \). To find how many such bombs are equivalent to the energy calculated in Step 1:\[ \text{Number of bombs} = \frac{1.44 \times 10^{17} \text{ J}}{2 \times 10^{16} \text{ J/bomb}} = 7.2 \text{ bombs} \]
04

- Calculate Mass Equivalent per Bomb Explosion

Using Einstein's mass-energy equivalence formula, rearrange to solve for mass:\[ m = \frac{E}{c^2} \]Here, \( E = 2 \times 10^{16} \text{ J} \) for a single bomb, and \( c = 3 \times 10^8 \text{ m/s} \):\[ m = \frac{2 \times 10^{16} \text{ J}}{(3 \times 10^8 \text{ m/s})^2} = \frac{2 \times 10^{16} \text{ J}}{9 \times 10^{16} \text{ m}^2/\text{s}^2} = \frac{2}{9} \text{ kg} = 0.22 \text{ kg} \]

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Key Concepts

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

Einstein's formula
Einstein's formula, represented by the equation \[E = mc^2\], explains the relationship between mass and energy. This iconic equation reveals that mass can be converted into a significant amount of energy. The equation consists of the following parameters: \[E\] - Energy, \[m\] - Mass, and \[c\] - the speed of light in a vacuum (approximately \[3 \times 10^8 \] meters per second).
The equation implies that even a small amount of mass can be converted into a vast amount of energy because the speed of light squared is a very large number. In physics, this principle is fundamental to understanding nuclear reactions and processes in astrophysics.
Hydrogen bomb energy
The hydrogen bomb, also known as a thermonuclear bomb, releases a massive amount of energy through nuclear fusion. In our example, the energy of a 5-megaton hydrogen bomb is \[2 \times 10^{16} \text{ J}\].
To comprehend the scale: \[\bullet \text{A single bomb yields more energy than the total energy used by some nations in a year.}\]
In the exercise, we calculate how many 5-megaton hydrogen bombs are needed to match the energy that would be produced by converting the mass of a textbook into energy.
Such calculations highlight the devastating potential of hydrogen bombs and underscore the importance of understanding nuclear energy.
Mass to energy conversion
Mass to energy conversion leverages Einstein's formula to transform mass into energy. This principle is essential in nuclear physics.
In the exercise, converting a 1.6 kg textbook to energy produces \[1.44 \times 10^{17} \text{ J}\]. By comparing this to the energy of a hydrogen bomb, we see it is equivalent to roughly 7.2 hydrogen bombs.
To make this conversion, we use: \[mc^2 \]
This equation is crucial for calculating how much energy can be derived from a particular mass and is underlined in various technologies, including nuclear reactors and particle accelerators.
Nuclear fusion in the Sun
Nuclear fusion is the process powering the Sun. It involves combining lighter atomic nuclei (like hydrogen) to form a heavier nucleus (like helium), releasing energy in the process.
In the Sun, hydrogen nuclei fuse under extreme pressure and temperature to form helium, releasing energy that powers sunlight.
This fusion process is fundamentally similar to what occurs in hydrogen bombs but on a much larger and controlled scale.
Understanding this mechanism helps scientists develop technologies for controlled nuclear fusion, which could provide nearly infinite energy with minimal environmental impact.

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Most popular questions from this chapter

The Sun shines by converting mass into energy according to \(E=m c^{2} .\) Show that if the Sun produces \(3.85 \times 10^{26} \mathrm{J}\) of energy per second, it must convert 4.3 million metric tons \(\left(4.3 \times 10^{\circ} \mathrm{kg}\right)\) of mass per second into energy.

The solar neutrino problem pointed to a fundamental gap in our knowledge of a. nuclear fusion. b. neutrinos. c. hydrostatic equilibrium. d. magnetic fields.

Place in order the following steps in the fusion of hydrogen into helium. If two or more steps happen simultaneously, use an equals sign ( =). a. A positron is emitted. b. One gamma ray is emitted. c. Two hydrogen nuclei are emitted. d. Two \(^{3}\) He collide and become \(^{4}\) He. e. Two hydrogen nuclei collide and become \(^{2} \mathrm{H}\). f. Two gamma rays are emitted. g. A neutrino is emitted. h. One deuterium nucleus and one hydrogen nucleus collide and become \(^{3}\) He.

The Sun rotates once every 25 days relative to the stars. The Sun rotates once every 27 days relative to Earth. Why are these two numbers different? a. The stars are farther away. b. Earth is smaller. c. Earth moves in its orbit during this time. d. The Sun moves relative to the stars.

The solar wind pushes on the magnetosphere of Earth, changing its shape, because a. the solar wind is so dense. b. the magnetosphere is so weak. c. the solar wind contains charged particles. d. the solar wind is so fast.
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