Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The starship \(Enterprise\), of television and movie fame, is powered by combining matter and antimatter. If the entire 400-kg antimatter fuel supply of the \(Enterprise\) combines with matter, how much energy is released? How does this compare to the U.S. yearly energy use, which is roughly \(1.0 \times 10^{20}\) J ?

Short Answer

Expert verified
The energy released is \( 7.2 \times 10^{19} \) J, which is 0.72 times the U.S. yearly energy use.

Step by step solution

01

Understand the Concept of Matter-Antimatter Annihilation

In matter-antimatter annihilation, when antimatter meets matter, they annihilate each other, converting their entire mass into energy according to Einstein's mass-energy equivalence principle, given by the equation \[ E = mc^2 \]. This means the released energy \( E \) can be found by multiplying the combined mass \( m \) by the speed of light squared \( c^2 \).
02

Calculate the Combined Mass

The problem states that the starship has a 400-kg antimatter fuel supply that completely annihilates with an equal amount (400 kg) of matter. Therefore, the total mass \( m \) getting converted into energy is \[ m = 400\, \text{kg} + 400\, \text{kg} = 800\, \text{kg}. \]
03

Apply Einstein's Equation

Insert the total mass and the speed of light into the equation \( E = mc^2 \). The speed of light \( c \) is approximately \( 3.0 \times 10^8 \) m/s, so the energy is \[ E = 800\, \text{kg} \times (3.0 \times 10^8\, \text{m/s})^2. \]
04

Perform the Calculation

Calculate the energy released by computing: \[ E = 800 \times 9.0 \times 10^{16} = 7.2 \times 10^{19}\, J. \]
05

Compare with U.S. Yearly Energy Use

The energy released by the annihilation process is \( 7.2 \times 10^{19} \) J. Comparing this with the U.S. yearly energy use of \( 1.0 \times 10^{20} \) J shows that the energy from the matter-antimatter annihilation is slightly less — about \( 0.72 \) times — than the total energy used by the U.S. in a year.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Einstein's mass-energy equivalence
Einstein's mass-energy equivalence is a fascinating breakthrough that revolutionized our understanding of physics. It tells us that mass and energy are two forms of the same thing and can be converted into each other. This principle is encapsulated in the famous equation - \[ E = mc^2 \] - where:
  • \( E \) represents energy in joules (J)
  • \( m \) symbolizes mass in kilograms (kg)
  • \( c \) denotes the speed of light in meters per second (m/s), specifically \( 3.0 \times 10^8 \ \text{m/s} \)
This means if you convert any mass fully into energy, its energy output is the mass times the speed of light squared. In scenarios like matter-antimatter annihilation, both the matter and antimatter are completely transformed, unleashing massive amounts of energy.
Energy calculation
To find out how much energy the starship \(Enterprise\) can release, we need to apply Einstein's equation correctly. We begin by understanding we have both matter and antimatter each weighing 400 kg. When they combine, they form an 800 kg mass that undergoes annihilation. Therefore, our combined mass is - \[ m = 400 \, \text{kg} + 400 \, \text{kg} = 800 \, \text{kg} \].- Next, we plug this total mass into Einstein's formula:
  • \( E = mc^2 = 800 \, \text{kg} \times (3.0 \times 10^8 \, \text{m/s})^2 \)
The key point is calculating the speed of light squared, giving \( 9.0 \times 10^{16} \). Our final energy calculation turns into:
  • \( E = 800 \times 9.0 \times 10^{16} = 7.2 \times 10^{19} \, \text{J} \).
This calculation tells us that a huge amount of energy, \( 7.2 \times 10^{19} \ \text{J} \), could be unleashed through this process.
Comparison with energy consumption
To put this tremendous release of energy into perspective, compare it to the yearly energy consumption of the United States. The U.S. uses approximately \( 1.0 \times 10^{20} \) joules each year. Observing this, the energy from the matter-antimatter reaction is slightly less than what the entire country consumes annually.- Since the energy provided is \( 7.2 \times 10^{19} \ \text{J} \), dividing this by \( 1.0 \times 10^{20} \ \text{J} \) tells us how significant this energy output is.
  • The reaction provides about 72% of what the U.S. uses yearly.
Such a comparison highlights how even a relatively small amount of matter and antimatter can produce an enormous quantity of energy, showcasing the unparalleled potential for power if matter-antimatter reactions could be controlled for practical use.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\textbf{Radiation Therapy with \)\pi^-\( Mesons.}\) Beams of \(\pi^-\) mesons are used in radiation therapy for certain cancers. The energy comes from the complete decay of the \(\pi^-\) to stable particles. (a) Write out the complete decay of a \(\pi^-\) meson to stable particles. What are these particles? (b) How much energy is released from the complete decay of a single \(\pi^-\) meson to stable particles? (You can ignore the very small masses of the neutrinos.) (c) How many \(\pi^-\) mesons need to decay to give a dose of 50.0 Gy to 10.0 g of tissue? (d) What would be the equivalent dose in part (c) in Sv and in rem? Consult Table 43.3 and use the largest appropriate RBE for the particles involved in this decay.

Suppose that positron-electron annihilations occur on the line 3 cm from the center of the line connecting two detectors. Will the resultant photons be counted as having arrived at these detectors simultaneously? (a) No, because the time difference between their arrivals is 100 ms; (b) no, because the time difference is 200 ms; (c) yes, because the time difference is 0.1 ns; (d) yes, because the time difference is 0.2 ns.

In which of the following decays are the three lepton numbers conserved? In each case, explain your reasoning. (a) \(\mu^-\rightarrow e^- + \nu_e + \overline{\nu}_\mu\); (b) \(\tau^-\rightarrow e^- + \overline{\nu}_e + \overline {\nu} _\tau\); (c) \(\pi^+ \rightarrow e^+ + \gamma\); (d) \(n \rightarrow p + e^- + \overline{\nu}_e\).

The magnetic field in a cyclotron that accelerates protons is 1.70 T. (a) How many times per second should the potential across the dees reverse? (This is twice the frequency of the circulating protons.) (b) The maximum radius of the cyclotron is 0.250 m. What is the maximum speed of the proton? (c) Through what potential difference must the proton be accelerated from rest to give it the speed that you calculated in part (b)?

Calculate the threshold kinetic energy for the reaction \(p + p \rightarrow p + p + K^+ + K^-\) if a proton beam is incident on a stationary proton target.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free