Is it possible to calculate the acceleration of an alpha particle emitted by a radionuclide, for example Po-210? I suppose the alpha particle appears, after tunneling out of the nucleus, at rest at a small distance $r_0$ from the center. The repulsive Coulomb force quickly accelerates the alpha particle to a high speed. In the case of polonium-210 the final kinetic energy, $E$, is 5.4 MeV. The distance r0 can be calculated from $E$:

$$ E = \frac{f q Q}{r_0} \Rightarrow r_0 = \frac{f q Q}{E} = 4⋅10^{-14} m $$

The initial acceleration at $r = r_0$, due to the repulsive Coulomb force, is: $$ a_0 = \frac{F}{m} = \frac{f q Q}{{r_0}^2 m} = \frac{f q Q}{\left(\frac{f q Q}{E}\right)^2 m} = \frac{E^2}{f q Q m} = 3⋅10^{27} \frac{m}{s^2} $$

Is this a valid idea? I ignored Heisenberg's uncertainty relation, because at $r = 1.3⋅r_0$ the acceleration is almost equally high, while the velocity of the particle, $ v = \sqrt{\frac{2}{m} (E-\frac{f q Q}{r}) } $, is 8⋅106 m/s, and the De Broglie wavelength is smaller than $r$.

If the idea is invalid because the alpha particle is a probablility wave, then at what distance does it become sufficiently particle-like to allow the calculation of acceleration?

EDIT - I meant the acceleration of the center of mass of the alpha particle, a = (1/m) dp/dt. I assumed the recoil of the daughter nucleus was the instantaneous measurement. Both the alpha particle and the nucleus are not point particles, but composite systems. Would perhaps the Rutherford variant be more classic and simple, and less affected by spherical symmetry assumptions, than the initial emission of the alpha particle? Rutherford fired alpha particles from a radioactive source to a gold target, and some of the alpha particles were backscattered. So the Rutherford variant of my question is the alpha-particle emitted by Po-210-decay being backscattered by another Po-210-nucleus. According to the formulas, the same trajectories and the same x0 and a0 seem to apply, approximately, without any spherical symmetry boundary condition. Is the maximum acceleration a0 = E2 / (fqQm) approximately right in the Rutherford variant of my question?


1 Answer 1


The calculation looks correct under the given assumptions. But the assumptions don't really match reality, acceleration is a quantity that no longer works on the atomic level.

A few notes on specific problems with your assumptions:

  1. Quantum mechanics is highly relevant for the process, so talking about acceleration is a bit of a stretch. The $\alpha$-particle leaves the nucleus as some probability wave, not as a point-particle that's accelerated. Depending where you make the semi-classical cut or assume measurement, you could describe the situation by an exponentially weakening outgoing probability wave of the $\alpha$-particle.

  2. Calling accelerations of nuclear particles as "occuring in everyday life" is a bit of a stretch. The acceleration of quantum particles can't be connected to everyday experience. E.g. the particle would have to radiate classical bremsstrahlung due to the acceleration if the situation could be described classically, but the real emissions behaviour deviates. (I could actually find a dissertation about bremsstrahlung emitted during $\alpha$-decay exactly for the case of $\,{}^{210}\text{Po}$ https://archiv.ub.uni-heidelberg.de/volltextserver/9535/. The theoretical part also discussion the quasi-classical approximations.)

I had a point about the initial kinetic energy of the $\alpha$-particle after tunnelling, but the semi-classical models presented in the linked thesis set this to zero – so I was wrong there.

(I thought about the wave-functions of incident monochromatic waves tunnelling through sharp barriers, those have kinetic energy instantly after leaving the barrier in the sense of having a defined wave-vector, but the case is different for a continuously varying potential).

  • $\begingroup$ The alpha particle is obviously a composite particle, not a point particle. In my mind acceleration of a composite particle applies to its center of mass. I am not well versed in quantum physics, but I don't see why a probability wave does not have a center of mass. In my mind, a = (1/m) dp/dt. So saying that a probability wave does not have an acceleration seems to imply that momentum is not defined for a probability wave. Is that true? $\endgroup$
    – jkien
    Commented Jul 4, 2021 at 10:44
  • $\begingroup$ You can separate the center of mass in the coordinates of the wave-function describing a composite particle (and then, unter certain circumstances), separate off a Schrödinger equation for that. But the wave function for the emitted alpha-particle is rotation symmetric, so the acceleration defined in terms of the momentum average is zero. This only changes when the position or momentum of the particle is measured, and "the wave function collapses" to the measured state. Before that, it is a spherical wave. $\endgroup$ Commented Jul 4, 2021 at 11:26
  • $\begingroup$ Wouldn't the daughter nucleus recoil instantaneously when the alpha particle is emitted, and wouldn't that count as the measurement of momentum? Or is the recoil actually delayed until the alpha particle is measured by a remote detector, at a later time? $\endgroup$
    – jkien
    Commented Jul 4, 2021 at 19:39
  • $\begingroup$ The recoil is not delayed. But it is also non-classical (that is, the wave function of the centre of mass of the daughter nucleus is also a spherical wave – and when you consider the full state the momenta are entangled, so that if the recoil momentum or the alpha momentum is measured, the other one is fixed in subsequent measurements which ensures conservation of momentum). $\endgroup$ Commented Jul 6, 2021 at 16:07

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