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Chapter 2: Problem 29

What is the speed of a particle that is observed to have momentum \(500 \mathrm{MeV} / \mathrm{c}\) and energy \(1746 \mathrm{MeV} ?\) What is the particle's mass (in \(\mathrm{MeV} / c^{2}\) )?

Short Answer

Expert verified
The speed of the particle is approximately 0.286c, and the particle's rest mass is roughly 938 MeV/c虏.

Step by step solution

01

Convert units if necessary

Ensure that the units are consistent and suitable for the calculations. In this case, the momentum is already given in MeV/c, which is appropriate for use in relativistic equations, and the energy is given in MeV. Thus, no unit conversion is necessary.
02

Calculate Speed Using Energy and Momentum

Use the relativistic relation between energy (E), momentum (p), and speed (v): \( E^{2} = (pc)^{2} + (m_{0}c^{2})^{2} \), where \( m_{0} \) is the rest mass and c is the speed of light in vacuum. Because we want to find the speed, we can rearrange the formula to solve for v: divide both sides by \( E^{2} \) to get \( 1 = (\frac{p}{E})^{2}c^{2} + (\frac{m_{0}c^{2}}{E})^{2} \); since we're interested in the term with velocity, isolate that term to get \( (\frac{p}{E})^{2}c^{2} = 1 - (\frac{m_{0}c^{2}}{E})^{2} \), then find v by taking the square root and dividing by c: \( v = \frac{p}{E}c \).
03

Calculate the Speed of the Particle

Insert the given values of momentum (500 MeV/c) and energy (1746 MeV) into the equation from Step 2 to find v. So \( v = \frac{500}{1746}c \). After calculating this, you should find the speed of the particle in terms of c.
04

Calculate the Particle's Rest Mass

From the same relativistic relation as in Step 2, solve for the rest mass \( m_{0} \): \( m_{0} = \frac{\sqrt{E^{2} - (pc)^{2}}}{c^{2}} \). Plug in the given values for E and p and perform the calculation to find the rest mass \( m_{0} \) in MeV/c虏.

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Key Concepts

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

Relativistic Energy-Momentum Relationship
In relativistic particle physics, the energy-momentum relationship is a fundamental concept explaining the connection between a particle's energy, momentum, and rest mass. This relationship is encapsulated by the equation \( E^{2} = (pc)^{2} + (m_{0}c^{2})^{2} \), where \(E\) is the total energy of the particle, \(p\) its momentum, \(m_{0}\) the rest mass, and \(c\) the speed of light in a vacuum.

The equation reveals that even when a particle is at rest (momentum \(p=0\)), it still has energy鈥攖he rest energy, given by \(E = m_{0}c^{2}\). This is why even massless particles, like photons, that travel at the speed of light still carry momentum. When dealing with problems in relativistic dynamics, it's essential to ensure that units are consistent, typically expressed in electron volts (eV) for energy and natural units (where \(c=1\)) for speed in mathematical calculations.
Particle Speed Calculation
To calculate the speed of a relativistic particle, a rearrangement of the energy-momentum relationship is required. Initially isolating the momentum term, we derive \( v = \frac{p}{E}c \) by assuming that the mass of the particle \(m_{0}\) does not change with speed.

In the given problem, the particle's momentum is specified as \(500 \mathrm{MeV}/c\) and its total energy as \(1746 \mathrm{MeV}\). Plugging these values into the speed equation, the particle's velocity in terms of the speed of light, \(c\), can be determined. It's critical for students to remember that the speeds obtained in such calculations are often significant fractions of \(c\), reinforcing the necessity of relativistic over classical mechanics in these scenarios.
Rest Mass Determination
The determination of a particle's rest mass \(m_{0}\) is conducted through the manipulation of the energy-momentum relationship. By isolating the rest mass term from the equation \( m_{0} = \frac{\sqrt{E^{2} - (pc)^{2}}}{c^{2}} \), we can find the intrinsic mass of the particle, which is independent of its motion.

In practice, when provided with the energy and momentum, as seen in our example, we solve for \(m_{0}\) by inserting the given values and calculating \( \sqrt{E^{2} - (pc)^{2}} \) followed by division by \( c^{2} \). This calculation yields the mass in the unit \(\mathrm{MeV}/c^{2}\), a common unit used in particle physics to express rest mass, emphasizing the mass-energy equivalence.

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Most popular questions from this chapter

An unstable particle of mass \(M\) decays into two identical particles, each of mass \(m\). Obtain an expression for the velocities of the two decay particles in the lab frame ( \(a\) ) if \(M\) is at rest in the lab and \((b)\) if \(M\) has total energy \(4 m c^{2}\) when it decays and the decay particles move along the direction of \(M\).

A proton with rest energy of \(938 \mathrm{MeV}\) has a total energy of \(1400 \mathrm{MeV}\). ( \(a\) ) What is its speed? (b) What is its momentum?

A pion spontaneously decays into a muon and a muon antineutrino according to (among other processes) \(\pi^{-} \rightarrow \mu^{-}+\bar{v}_{\mu}\). Recent experimental evidence indicates that the mass \(m\) of the \(\bar{v}_{\mu}\) is no larger than about \(190 \mathrm{keV} / \mathrm{c}^{2}\) and may be as small as zero. Assuming that the pion decays at rest in the laboratory, compute the energies and momenta of the muon and muon antineutrino \((a)\) if the mass of the antineutrino were zero and \((b)\) if its mass were \(190 \mathrm{keV} / \mathrm{c}^{2}\). The mass of the pion is \(139.56755 \mathrm{MeV} / \mathrm{c}^{2}\) and the mass of the muon is \(105.65839 \mathrm{MeV} / \mathrm{c}^{2}\). (See Chapters 11 and 12 for more on the neutrino mass.)

Compute the rest energy of \(1 \mathrm{~g}\) of dirt. \((b)\) If you could convert this energy entirely into electrical energy and sell it for 10 cents per kilowatt-hour, how much money would you get? \((c)\) If you could power a \(100 \mathrm{~W}\) lightbulb with the energy, for how long could you keep the bulb lit?

An elementary particle of mass \(M\) completely absorbs a photon, after which its mass is \(1.01 M\). ( \(a\) ) What was the energy of the incoming photon? ( \(b\) ) Why is that energy greater than \(0.01 M c^{2} ?\)
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