Problem — Rest energy of the electron

Problem

Calculate the rest energy of an electron using $E_0 = m c^2$. Express the result in joules and in MeV.

Solution

Data. Electron mass $m = 9{,}11\cdot 10^{-31}$ kg; speed of light $c = 3{,}0\cdot 10^{8}$ m/s.

Rest energy in joules.

$E_0 = m c^2 = (9{,}11\cdot 10^{-31}\;\text{kg})(3{,}0\cdot 10^{8}\;\text{m/s})^2 = 9{,}11\cdot 10^{-31}\times 9{,}0\cdot 10^{16}\;\text{J}$

$\ev{E_0 \approx 8{,}2\cdot 10^{-14}\;\text{J}}$

Conversion to MeV. Recalling that $1\;\text{eV} = 1{,}60\cdot 10^{-19}$ J, so $1\;\text{MeV} = 1{,}60\cdot 10^{-13}$ J:

$E_0 = \frac{8{,}2\cdot 10^{-14}\;\text{J}}{1{,}60\cdot 10^{-13}\;\text{J/MeV}}$

$\ev{E_0 = 8{,}2\cdot 10^{-14}\;\text{J} = 0{,}511\;\text{MeV}}$

The value $0{,}511$ MeV is the electron’s characteristic rest energy, often used as a natural unit in particle physics.

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Topics: Relatività ristretta Concepts: Equivalenza massa-energia Skills: Impostazione simbolica