Electron spiraling into nucleus

Question

Any idea why it is said that, if electron is a particle it'll spiral into the nucleus and disintegrate in 10^-14 seconds?

https://youtu.be/2wF_CVuWyEg?t=123 to https://youtu.be/2wF_CVuWyEg?t=142

Answer 1

As per @probably_someone's answer, if electrons are point charge particles, they radiate energy when they are accelerated. If an electron orbit's a nucleus, the energy radiated per unit of time is given by larmor's formula:

$$P=\frac{2}{3}\frac{e^2a^2}{4\pi\epsilon_0c^3}$$

For a single electron orbiting a nucleus of charge $Z$, the acceleration is given by Newton's formula: $$ma = m\frac{v^2}{r}=\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2}$$ $$\rightarrow v=\sqrt{\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{mr}}$$

where $r$ is the distance between the electron and the nucleus. Knowing that $P=-\frac{dE}{dt}$ where $E$ is the electron's total energy. Using the above formula we get:

$$E=\frac{1}{2}mv^2-\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r}=-\frac{1}{8\pi\epsilon_0}\frac{Ze^2}{r}$$

Thus,

$$P=-\frac{dE}{dt}=-\frac{1}{8\pi\epsilon_0}\frac{Ze^2}{r^2}\frac{dr}{dt}=\frac{2}{3}\frac{e^2a^2}{4\pi\epsilon_0c^3}=\frac{2}{3}\frac{e^2}{4\pi\epsilon_0c^3}\left(\frac{1}{4\pi\epsilon_0 }\frac{Ze^2}{mr^2}\right)^2$$ $$\rightarrow \frac{dr}{dt}=-\frac{Ze^4}{12c^3\pi^2\epsilon_0^2m^2r^2}$$

Supposing that the electron start at the bohr radius $r_0$ for an atom of charge $Z$, $r_0=\frac{4\pi\epsilon_0\hbar^2}{mZe^2}$ and integrating the differential equation, we get the time $t$ when the electron "crashes" into the nucleus:

$$t=\int_0^tdt' = \int_{r_0}^0-\frac{12c^3\pi^2\epsilon_0^2m^2r^2}{Ze^4}dr=\frac{4c^3\pi^2\epsilon_0^2m^2r_0^3}{Ze^4}\sim 10^{-11}s$$

(for $Z=1$) which is very quick!