What is the rest energy of an electron, given its mass is \(9.11 \times 10^{-31} \mathrm{kg} ?\) Give your answer in joules and \(\mathrm{MeV}\)

Short Answer

Expert verified
The rest energy of the electron is \(8.187 \times 10^{-14} \mathrm{J}\) or \(0.511 \mathrm{MeV}\).

Step by step solution

01

Write down the given values

We are given the mass of the electron: \(m = 9.11 \times 10^{-31} \mathrm{kg}\) and the speed of light in a vacuum: \(c = 2.998 \times 10^8 \mathrm{m/s}\).
02

Calculate the rest energy in joules

We will use Einstein's energy-mass relationship to calculate the rest energy: \(E = mc^2\) Plug in the given values: \(E = (9.11 \times 10^{-31} \mathrm{kg}) (2.998 \times 10^8 \mathrm{m/s})^2\) Calculate the energy: \(E = 8.187 \times 10^{-14} \mathrm{J}\)
03

Convert the rest energy from joules to MeV

To convert the energy from joules to MeV, we can use the conversion factor: \(1 \mathrm{MeV} = 1.602 \times 10^{-13} \mathrm{J}\) Divide the energy in joules by the conversion factor: \(E = \frac{8.187 \times 10^{-14} \mathrm{J}}{1.602 \times 10^{-13} \mathrm{J/MeV}}\) Calculate the energy: \(E = 0.511 \mathrm{MeV}\) So, the rest energy of the electron is \(8.187 \times 10^{-14} \mathrm{J}\) or \(0.511 \mathrm{MeV}\).

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