Question

An electron's rest mass is \(0.511 \mathrm{MeV} / c^{2}\) a) How fast must an electron be moving if its energy is to be 10 times its rest energy? b) What is the momentum of the electron at this speed?

Short answer

Answer: a) The speed of the electron when its energy is 10 times its rest energy is given by \(v = c\sqrt{1 - \frac{1}{10^2}}\). b) The momentum of the electron at this speed is given by \(p = 10m \cdot c\sqrt{1 - \frac{1}{10^2}}\).

Step-by-step solution

Write down the relativistic total energy equation

The relativistic total energy equation is given by \(E = \gamma mc^2\), where \(E\) is the total energy, \(\gamma\) is the Lorentz factor, \(m\) is the rest mass, and \(c\) is the speed of light. The Lorentz factor is defined as \(\gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}}\), where \(v\) is the speed of the electron.

Find the condition for total energy to be 10 times rest energy

We are given that the total energy of the electron should be 10 times its rest energy. Mathematically, this can be written as \(E = 10mc^2\). Substituting the relativistic total energy equation, we get \(\gamma mc^2 = 10mc^2\).

Solve for the speed v

To find the speed \(v\), we can divide both sides of the equation in Step 2 by \(mc^2\) and then isolate \(\gamma\). We're looking for the speed \(v\) that makes \(\gamma = 10\), so we can write the equation as \(\frac{1}{\sqrt{1 - (\frac{v}{c})^2}} = 10\). Squaring both sides and solving for \(\frac{v}{c}\), we get \(\frac{v}{c} = \sqrt{1 - \frac{1}{10^2}}\). From this, we can solve for \(v\), getting \(v = c\sqrt{1 - \frac{1}{10^2}}\).

Answer to part a

The speed \(v\) of the electron, such that its energy is 10 times its rest energy, is given by \(v = c\sqrt{1 - \frac{1}{10^2}}\).

Write down the relativistic momentum equation

The relativistic momentum equation is given by \(p = \gamma mv\), where \(p\) is the momentum, \(\gamma\) is the Lorentz factor, \(m\) is the rest mass, and \(v\) is the speed of the electron.

Substitute the speed from part a and find the momentum

We have already found the speed \(v\) in part a. Now we substitute it into the relativistic momentum equation. We get \(p = 10m \cdot c\sqrt{1 - \frac{1}{10^2}}\).

Answer to part b

The momentum of the electron at the speed found in part a is given by \(p = 10m \cdot c\sqrt{1 - \frac{1}{10^2}}\).