Amateur Topologist

Everything but topology.

Electrons are not like planets

One of my first posts on this blog was about the stability of the atomic nucleus; given that it consists of a bunch of positive/neutral charges clumped together, why doesn’t it fly apart? The answer involves the strong force, which is strong at atomic distances but miniscule at inter-nuclear distances; on distances comparable to that of an atomic nucleus, it’s strong enough to overcome the electromagnetic repulsion. But there’s another question, and it involves the model of orbiting electrons.

Classically (meaning without quantum mechanics), the electron is pictured as a pointlike particle orbiting the nucleus. For simplicitly, we’ll look at the hydrogen atom. The electron orbits at the Bohr radius , a_0=5.3 \times 10^{-13} cm, which can’t really be derived in an easy way from theory; we’ll just take it as a given. So the acceleration the electron undergoes is q^2 / m_e r^2 (using cgs units to avoid factors of 1/4\pi \epsilon_0 and such everywhere), using good old F=ma.

However, there is a problem: any accelerating charge radiates energy. The reason for this is, roughly speaking, an accelerating charge has more energy in its electromagnetic field, so you therefore have to expend more energy to accelerate it; since an atom is a closed system, there can be no energy source, so it ‘extracts’ it by spiraling inwards. The derivation of formula is complicated, but it turns out that a particle of charge q accelerating at a rate a will radiate energy at a rate

P=\frac{2 q^2 a^2}{3 c^3}=\frac{2q^6}{3m_e^2r^4c^3}

(the second equation what we get when we plug in our value for acceleration).

So what’s the energy of an electron orbiting its atom? The kinetic energy is equal to \frac{1}{2}mv^2=\frac{1}{2}rF=q^2/2r, and the potential energy is equal to $-q^2/r$, so the total energy is just

E=-q^2/2r

Considering both energy and radiated power as functions of time, we get dE/dt=-P; but since the only variable that can change is the radius, we then get the following differential equation for the radius:

r'=-\frac{4q^4}{3m^2c^3r^2}

The method of solving this equation isn’t important; I used Mathematica. What is important is the solution:

r=\left(a_0^3-\frac{4q^4t}{c^3m^2}\right)^{1/3}

where a_0 is the Bohr radius I mentioned earlier. This will obviously become zero at t_0=\frac{a_0^3m_e^2c^3}{4q^4}; plugging in the cgs values for these constants, we get that t_0=3.11 \times 10^{-11} seconds. So according to the Bohr model, a hydrogen atom decays in less than the time it takes light to move a centimeter.

So what’s the answer? One is to use the Bohr model, which requires a minimum energy; the electron cannot fall farther into an energy well. This model works to a certain extent, but fails with more complicated atoms; not only that, but it predicts that the hydrogen atom has a minimum nonzero angular momentum, which is not the case. But a full treatment of the failings of the Bohr model goes beyond what I know.

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