Select The Correct Answer.If There Is A Loss Of $4.36 \times 10^{-5} \, \text{g}$ Of Mass In A Nuclear Reaction, How Many KJ Of Energy Would Be Released? Recall That $c = 3 \times 10^8 \, \text{m/s}$.A. $1.45 \times

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Introduction

The concept of mass-energy equivalence, as described by Albert Einstein's famous equation E=mc^2, is a fundamental principle in physics that relates the energy of a system to its mass. In the context of nuclear reactions, this principle is particularly relevant, as it allows us to calculate the energy released or absorbed during a reaction. In this article, we will explore how to apply the mass-energy equivalence principle to a specific scenario involving a loss of mass in a nuclear reaction.

The Mass-Energy Equivalence Equation

The mass-energy equivalence equation is given by:

E = mc^2

where E is the energy of the system, m is the mass of the system, and c is the speed of light in vacuum. This equation shows that energy (E) is directly proportional to mass (m) and the speed of light (c) squared.

Calculating Energy Released in a Nuclear Reaction

Let's consider a scenario where there is a loss of mass of $4.36 \times 10^{-5} , \text{g}$ in a nuclear reaction. We want to calculate the energy released during this reaction. To do this, we need to convert the mass loss from grams to kilograms, as the speed of light is typically expressed in meters per second.

m=4.36×10−5 g=4.36×10−8 kgm = 4.36 \times 10^{-5} \, \text{g} = 4.36 \times 10^{-8} \, \text{kg}

Now, we can plug in the values into the mass-energy equivalence equation:

E = mc^2

E=(4.36×10−8 kg)×(3×108 m/s)2E = (4.36 \times 10^{-8} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2

E = (4.36 \times 10^{-8} \, \text{kg}) \times (9 \times 10^{16} \, \text{m^2/s^2})

E=3.924×10−1 JE = 3.924 \times 10^{-1} \, \text{J}

To convert this energy from joules to kilojoules, we can divide by 1000:

E=3.924×10−1 J1000E = \frac{3.924 \times 10^{-1} \, \text{J}}{1000}

E=3.924×10−4 kJE = 3.924 \times 10^{-4} \, \text{kJ}

Conclusion

In conclusion, we have calculated the energy released during a nuclear reaction involving a loss of mass of $4.36 \times 10^{-5} , \text{g}$. Using the mass-energy equivalence equation, we found that the energy released is approximately 3.924 x 10^-4 kJ.

References

  • Einstein, A. (1905). Does the Inertia of a Body Depend Upon Its Energy Content? Annalen der Physik, 18(13), 639-641.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Discussion

What are some real-world applications of the mass-energy equivalence principle? How does this principle relate to nuclear power plants and nuclear medicine?

Answers

A. 1.45×10−3 kJ1.45 \times 10^{-3} \, \text{kJ}

Introduction

In our previous article, we explored the concept of mass-energy equivalence and its application to nuclear reactions. We calculated the energy released during a nuclear reaction involving a loss of mass of $4.36 \times 10^{-5} , \text{g}$. In this article, we will address some common questions and concerns related to mass-energy equivalence and nuclear reactions.

Q: What is the mass-energy equivalence equation?

A: The mass-energy equivalence equation is given by:

E = mc^2

where E is the energy of the system, m is the mass of the system, and c is the speed of light in vacuum.

Q: What is the significance of the speed of light in the mass-energy equivalence equation?

A: The speed of light (c) is a fundamental constant in physics that relates the energy of a system to its mass. The speed of light is approximately 3 x 10^8 m/s.

Q: How does the mass-energy equivalence principle relate to nuclear power plants?

A: The mass-energy equivalence principle is a crucial concept in nuclear power plants, where nuclear reactions are used to generate electricity. During a nuclear reaction, a small amount of mass is converted into energy, which is then used to heat water and produce steam. The steam drives a turbine, which generates electricity.

Q: What are some real-world applications of the mass-energy equivalence principle?

A: Some real-world applications of the mass-energy equivalence principle include:

  • Nuclear power plants: As mentioned earlier, nuclear power plants use nuclear reactions to generate electricity.
  • Nuclear medicine: Nuclear medicine uses radioactive isotopes to diagnose and treat diseases.
  • Particle accelerators: Particle accelerators use high-energy particles to study the properties of subatomic particles.
  • Space exploration: The mass-energy equivalence principle is used in space exploration to calculate the energy required to propel spacecraft.

Q: Can the mass-energy equivalence principle be used to create a nuclear bomb?

A: Yes, the mass-energy equivalence principle can be used to create a nuclear bomb. A nuclear bomb works by releasing a large amount of energy through a nuclear reaction, which is then converted into a massive explosion.

Q: What are some limitations of the mass-energy equivalence principle?

A: Some limitations of the mass-energy equivalence principle include:

  • The principle only applies to systems where the energy is released or absorbed through a nuclear reaction.
  • The principle does not account for the energy released or absorbed through other processes, such as chemical reactions or thermal expansion.

Conclusion

In conclusion, the mass-energy equivalence principle is a fundamental concept in physics that relates the energy of a system to its mass. The principle has numerous real-world applications, including nuclear power plants, nuclear medicine, particle accelerators, and space exploration. However, the principle also has limitations, which must be taken into account when applying it to real-world scenarios.

References

  • Einstein, A. (1905). Does the Inertia of a Body Depend Upon Its Energy Content? Annalen der Physik, 18(13), 639-641.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Discussion

What are some potential applications of the mass-energy equivalence principle in the future? How can the principle be used to improve our understanding of the universe?