Size of an electron

Electricity and Magnetism Level 2

We assume that the electron is a classical particle and has the shape of a sphere with radius r e r_e . At the sphere's surface, the entire electrical charge q = − e q = -e would be distributed. How big should the diameter 2 r e 2 r_e of the electron be, so that the electrostatic energy corresponds to the electron mass?

You might need to use these physical constants: elemental charge e ≈ 1.6 ⋅ 1 0 − 19 C electron mass m e ≈ 9.1 ⋅ 1 0 − 31 kg speed of light c ≈ 3 ⋅ 1 0 8 m/s vacuum permittivity ε 0 ≈ 8.9 ⋅ 1 0 − 12 F/m . \begin{aligned} \text{elemental charge} & & e &\approx 1.6 \cdot 10^{-19} \,\text{C} \\ \text{electron mass} & & m_e &\approx 9.1 \cdot 10^{-31} \,\text{kg} \\ \text{speed of light} & & c &\approx 3 \cdot 10^8\text{ m/s} \\ \text{vacuum permittivity} & & \varepsilon_0 &\approx 8.9 \cdot 10^{-12} \text{ F/m}. \end{aligned}

Note: In reality, the electron is subject to the laws of quantum mechanics. The calculated radius of the electron is therefore not meaningful because it cannot be located exactly due to the uncertainty relation. In an atom, the electron appears smeared over a volume that is much larger than the sphere assumed here.

3 nanometers 3 picometers 3 femtometers 3 attometers

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Dec 11, 2017

The electric field of the charged sphere equals ∣ E ⃗ ∣ = { e 4 π ε 0 1 r 2 r > r e 0 r ≤ r e |\vec E| = \begin{cases} \dfrac{e}{4 \pi \varepsilon_0} \dfrac{1}{r^2} & r > r_e \\ 0 & r \leq r_e \end{cases} Therefore, we get a corresponding field energy W el = ε 0 2 ∫ ∣ E ⃗ ∣ 2 d V = 2 π ε 0 ∫ 0 ∞ ∣ E ⃗ ∣ 2 r 2 d r = e 2 8 π ε 0 ∫ r e ∞ 1 r 2 d r = e 2 8 π ε 0 r e \begin{aligned} W_\text{el} &= \frac{\varepsilon_0}{2} \int |\vec E|^2 dV = 2 \pi \varepsilon_0 \int_0^\infty |\vec E|^2 r^2 dr \\ &= \frac{e^2}{8 \pi \varepsilon_0} \int_{r_e}^\infty \frac{1}{r^2} dr = \frac{e^2}{8 \pi \varepsilon_0 r_e} \end{aligned} Due to the mass–energy equivalence W el = m e c 2 W_\text{el} = m_e c^2 we get an classical electron radius of r e = e 2 8 π ε 0 m e c 2 ≈ 1.4 ⋅ 1 0 − 15 = 1.4 fm r_e = \frac{e^2}{8 \pi \varepsilon_0 m_e c^2} \approx 1.4 \cdot 10^{-15} = 1.4\,\text{fm} Other charge distributions lead to different results for the electron radius. However, the order of magnitude remains the same. The CODATA definition of the electron radius is twice the size r e = e 2 4 π ε 0 m e c 2 ≈ 3 fm r_e = \frac{e^2}{4 \pi \varepsilon_0 m_e c^2} \approx 3\,\text{fm}

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