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Chapter 26: Problem 49

An observer in the laboratory finds that an electron's total energy is $5.0 \mathrm{mc}^{2} .$ What is the magnitude of the electron's momentum (as a multiple of \(m c\) ), as observed in the laboratory?

Short Answer

Expert verified
Answer: The magnitude of the electron's momentum is \(\sqrt{24} mc\).

Step by step solution

01

Write the equation for the total energy of the electron

The total energy of the electron is given by the equation: \(E^2 = (mc^2)^2 + (pc)^2\) where \(E\) represents the energy of the electron, \(m\) is the mass of the electron, \(c\) represents the speed of light, and \(p\) represents the magnitude of the electron's momentum.
02

Rearrange the equation to solve for the magnitude of the momentum

To find the magnitude of the momentum, we need to rearrange the equation above to solve for \(p\). We can do this by subtracting \((mc^2)^2\) from both sides of the equation, then taking the square root: \((pc)^2 = E^2 - (mc^2)^2\) \(p = \frac{\sqrt{E^2 - (mc^2)^2}}{c}\)
03

Plug in the given values and compute the answer

Now we can plug in the given values to find the magnitude of the electron's momentum, as a multiple of \(mc\). We know \(E = 5.0\,\mathrm{mc}^2\), so: \(p = \frac{\sqrt{(5.0\,\mathrm{mc}^2)^2 - (mc^2)^2}}{c}\) \(p = \frac{\sqrt{(25 mc^2)^2 - (mc^2)^2}}{c}\) \(p = \frac{\sqrt{24\,(mc^2)^2}}{c}\) \(p = \sqrt{24} mc\) So, the magnitude of the electron's momentum is \(\sqrt{24} mc\), as observed in the laboratory.

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Most popular questions from this chapter

Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
An electron has a total energy of 6.5 MeV. What is its momentum (in MeV/c)?
Show that Eq. \((26-11)\) reduces to the nonrelativistic relationship between momentum and kinetic energy, \(K \approx p^{2} /(2 m),\) for \(K<E_{0}\).
A particle decays in flight into two pions, each having a rest energy of 140.0 MeV. The pions travel at right angles to each other with equal speeds of \(0.900 c .\) Find (a) the momentum magnitude of the original particle, (b) its kinetic energy, and (c) its mass in units of \(\mathrm{MeV} / c^{2}\).
The mean (average) lifetime of a muon in its rest frame is $2.2 \mu \mathrm{s}\(. A beam of muons is moving through the lab with speed \)0.994 c .$ How far on average does a muon travel through the lab before it decays?
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