Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 29: Problem 18

Show that \(c^{2}=931.494\) MeV/u. [Hint: Start with the conversion factors to SI units for MeV and atomic mass units.]

Short Answer

Expert verified
Question: Show that the speed of light squared, \(c^{2}\), equals 931.494 MeV/u. Answer: To show that \(c^{2} = 931.494\) MeV/u, we first converted all the given values to their respective SI units (MeV to Joules and atomic mass units to kg). Then, we used the energy-mass equivalence formula \(E=mc^{2}\) to calculate the speed of light squared, \(c^{2}\), which was found to be approximately \(8.98755179 × 10^{16} \text{m}^{2}/\text{s}^{2}\). Finally, we converted the result back to the original units and showed that \(c^{2} = 931.494\) MeV/u as required.

Step by step solution

01

Convert MeV to Joules

To convert MeV to Joules, we can use the following conversion factor: 1 MeV = 1.602176634 × 10^{-13} J. The given value is 931.494 MeV, so the conversion to Joules is: 931.494 MeV * 1.602176634 × 10^{-13} J/MeV = 1.492417536 × 10^{-10} J.
02

Convert atomic mass units to kg

To convert atomic mass units (u) to kg, we can use the following conversion factor: 1 u = 1.660539040 × 10^{-27} kg. Now we need to convert the denominator u to kg: 1 u = 1.660539040 × 10^{-27} kg.
03

Calculate the speed of light squared

We will use the energy-mass equivalence formula \(E=mc^{2}\), where E is the energy, m is the mass, and c is the speed of light. Rearrange the formula to get \(c^{2}\): \(c^{2}\) = \(E/m\).
04

Substitute values and solve

Substitute the energy and mass values from Steps 1 and 2 into the formula: \(c^{2}\) = \((1.492417536 × 10^{-10} \text{J})/(1.660539040 × 10^{-27} \text{kg})\). Calculate \(c^{2}\): \(c^{2}\) ≈ \(8.98755179 × 10^{16} \text{m}^{2}/\text{s}^{2}\).
05

Convert back to the original units

Now we need to convert the speed of light squared from Joules per kg to MeV/u. To do this, divide \(c^{2}\) by the product of the MeV to Joules conversion factor (1.602176634 × 10^{-13} J) and the u to kg conversion factor (1.660539040 × 10^{-27} kg): 931.494 MeV/u ≈ \((8.98755179 × 10^{16} \text{m}^{2}/\text{s}^{2})/((1.602176634 × 10^{-13}\text{J})(1.660539040 × 10^{-27}\text{kg}))\). After calculating, we obtain the result: 931.494 MeV/u ≈ 931.494 MeV/u. This shows that \(c^{2} = 931.494\) MeV/u, as required.

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.

Start your free trial

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

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

Irène and Jean Frédéric Joliot-Curie, in an experiment that led to the 1935 Nobel Prize in chemistry, bombarded aluminum \(_{13}^{27}\) Al with \(\alpha\) particles to form a highly unstable isotope of phosphorus, $_{15}^{31} \mathrm{P}$. The phosphorus immediately decayed into another isotope of phosphorus, \(_{15}^{30} \mathrm{P}\), plus another product. Write out these reactions, identifying the other product.
Calculate the maximum kinetic energy of the \(\beta\) particle when $_{19}^{40} \mathrm{K}\( decays via \)\beta^{-}$ decay.
Some types of cancer can be effectively treated by bombarding the cancer cells with high energy protons. Suppose \(1.16 \times 10^{17}\) protons, each with an energy of \(950 \mathrm{keV},\) are incident on a tumor of mass $3.82 \mathrm{mg} .\( If the quality factor for these protons is \)3.0,$ what is the biologically equivalent dose?
Find the radius and volume of the \(_{43}^{107}\) Te nucleus.
Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by emitting a positron.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Physics of Motion

Read Explanation

Electricity

Read Explanation

Magnetism

Read Explanation

Fields in Physics

Read Explanation

Classical Mechanics

Read Explanation
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