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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

When an electron travels at \(0.60 c,\) what is its total energy in mega- electron-volts?
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