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

Write the symbol (in the form \(_{Z}^{A} \mathrm{X}\) ) for the nuclide that has 78 neutrons and 53 protons.
Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?
The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.
Show mathematically that $2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\( if and only if \)T_{1 / 2}=\tau \ln 2 .$ [Hint: Take the natural logarithm of each side. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Electricity

Read Explanation

Work Energy and Power

Read Explanation

Torque and Rotational Motion

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