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Chapter 37: Problem 35

A particle has rest mass \(6.64 \times 10^{-27} \mathrm{~kg}\) and momentum \(2.10 \times 10^{-18} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} .\) (a) What is the total energy (kinetic plus rest energy) of the particle? (b) What is the kinetic energy of the particle? (c) What is the ratio of the kinetic energy to the rest energy of the particle?

Short Answer

Expert verified
Every value must be calculated. Here are the formulas for each step: Total energy: \(E = \sqrt{(pc)^2 + (mc^2)^2}\), kinetic energy: \(K.E. = E - mc^2\), ratio: \( K.E./(mc^2)\).

Step by step solution

01

Compute the Total Energy

To compute the total energy, we first need to convert rest mass to energy using mass-energy equivalence. To do this, multiply the rest mass \(6.64 \times 10^{-27}\) kg by the speed of light squared \(c^2 = (3 \times 10^8)^2\) m^2/s^2. The total energy E can then be calculated using the energy-momentum formula \(E = \sqrt{(pc)^2 + (mc^2)^2}\), where p is the given momentum \(2.10 \times 10^{-18} \) kg路m/s.
02

Calculate the Kinetic Energy

With the total energy E calculated, the kinetic energy can be computed as the difference between the total energy and the rest energy \(mc^2\). So, subtract the rest energy from the total energy to find the kinetic energy.
03

Find the Ratio of Kinetic Energy to Rest Energy

Finally, the ratio of the kinetic energy to the rest energy is found by simply dividing the kinetic energy by the rest energy. Use the values calculated in the previous steps to perform this division and arrive at the ratio.

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Key Concepts

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

Total Energy of a Particle
Understanding the total energy of a particle is fundamental in physics, particularly in the realms of quantum mechanics and relativity. The total energy encapsulates the entirety of an object's potential to do work or cause a change and is a combination of its kinetic energy 鈥� the energy due to motion 鈥� and its rest energy, associated with its mass. The famed equation by Albert Einstein, E = mc^2, highlights this relationship where E is the total energy, m is the rest mass, and c is the speed of light.
In practice, to calculate the total energy, we first convert the rest mass into energy by multiplying with the square of the speed of light. Then, we account for the momentum using the energy-momentum relationship, yielding a more general formula for total energy: E = \( \sqrt{(pc)^2 + (mc^2)^2}\). This elegant formula allows physicists to calculate the total energy for particles traveling at high speeds, where relativistic effects are significant, changing our conventional notions of energy.
  • Rest mass energy: Given by the product of mass and the square of the speed of light.
  • Kinetic energy: The added energy from the particle's motion.
  • Total energy: A comprehensive measure implicating both rest and kinetic energy.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It can be expressed using the classical formula KE = \(\frac{1}{2}mv^2\), where KE stands for kinetic energy, m is the mass of the object, and v is its velocity. However, when objects approach the speed of light, or for particles of very small mass such as electrons or protons, this simple formula is no longer accurate, and relativistic effects must be considered.

For such cases, the kinetic energy is determined by subtracting the rest energy (again, mc^2) from the total energy. This is done after first calculating the total energy which considers both the particle's rest mass and its momentum. It's a way of quantifying how much energy has been imparted to an object to set it in motion, over and above its intrinsic energy due to mass 鈥� a crucial piece of information in fields like particle physics and astrophysics.
  • Classical kinetic energy: Calculated for slower speeds and larger objects.
  • Relativistic kinetic energy: Calculated for particles at high speeds, accounting for the energy contributed by both mass and velocity.
Energy-Momentum Formula
The energy-momentum formula is central to understanding the dynamics of particles, especially within the framework of special relativity. It's a formula that reveals the deeper connection between an object's mass, its energy, and its momentum. Expressed as E = \(\sqrt{(pc)^2 + (mc^2)^2}\), where E is the total energy, p the momentum, and c the speed of light, it serves as a bridge between classical and relativistic mechanics.

This formula shows that the total energy is not just dependent on mass but is a function of momentum as well. As such, even massless particles, like photons, carry energy due to their momentum. By applying the energy-momentum formula to the exercise's given momentum and rest mass, one can find the total energy of the particle, which includes both its rest energy and kinetic energy. The significance of the energy-momentum formula cannot be overstated as it provides a comprehensive understanding of the behavior of particles in high-speed contexts, laying the foundation for much of modern physics.
  • Connects energy with mass and momentum.
  • Applicable to particles with and without rest mass.
  • Foundational in both theoretical and applied physics.

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Most popular questions from this chapter

As a research scientist at a linear accelerator, you are studying an unstable particle. You measure its mean lifetime \(\Delta t\) as a function of the particle's speed relative to your laboratory equipment. You record the speed of the particle \(u\) as a fraction of the speed of light in vacuum \(c .\) The table gives the results of your measurements. $$ \begin{array}{l|lllllll} u / c & 0.70 & 0.80 & 0.85 & 0.88 & 0.90 & 0.92 & 0.94 \\ \hline \Delta t\left(10^{-8} \mathrm{~s}\right) & 3.57 & 4.41 & 5.02 & 5.47 & 6.05 & 6.58 & 7.62 \end{array} $$ (a) Your team leader suggests that if you plot your data as \((\Delta t)^{2}\) versus \(\left(1-u^{2} / c^{2}\right)^{-1},\) the data points will be fit well by a straight line. Construct this graph and verify the team leader's prediction. Use the best-fit straight line to your data to calculate the mean lifetime of the particle in its rest frame. (b) What is the speed of the particle relative to your lab equipment (expressed as \(u / c\) ) if the lifetime that you measure is four times its rest-frame lifetime?

A particle zips by us with a Lorentz factor of \(1.12 .\) Then another particle zips by us moving at twice the speed of the first particle. (a) What is the Lorentz factor of the second particle? (b) If the particles were moving with a speed much less than \(c,\) the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?

Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called spectroscopic binary stars. Figure \(\mathbf{P 3 7 . 6 8}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R .\) The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times 10^{14} \mathrm{~Hz}\) In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{~Hz}\) and \(4.568910 \times 10^{14} \mathrm{~Hz}\). Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f / f=u / c\) given in Section \(37.6 .\) ) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30} \mathrm{~kg} .\) Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{~m}\). (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of \(0.520 c\). If the radiation has a frequency of \(8.64 \times 10^{14} \mathrm{~Hz}\) in the rest frame of the star, what is the frequency measured by an observer on earth?

Two events are observed in a frame of reference \(S\) to occur at the same space point, the second occurring \(1.80 \mathrm{~s}\) after the first. In a frame \(S^{\prime}\) moving relative to \(S\), the second event is observed to occur \(2.15 \mathrm{~s}\) after the first. What is the difference between the positions of the two events as measured in \(S^{\prime} ?\)
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