Maximum acceleration of an alpha particle?

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SUMMARY

The maximum acceleration of an alpha particle backscattered by a heavy atom, such as polonium, is estimated to be on the order of 1⋅1027 m/s2 when considering a kinetic energy of 5.4 MeV. This estimation is derived from equating the kinetic energy to electrostatic potential energy, leading to the formula a0 = E2 / (f q Q m). The analysis incorporates quantum mechanical considerations, specifically the de Broglie wavelength, suggesting that the relevant distance for acceleration calculations should be at least 10 times this wavelength. The discussion highlights the transition from wave-like to particle-like behavior in alpha particles during interactions.

PREREQUISITES
  • Understanding of electrostatic potential energy and Coulomb's law
  • Familiarity with de Broglie wavelength and quantum mechanics
  • Knowledge of kinetic energy and its relation to particle acceleration
  • Basic principles of Rutherford backscattering
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  • Research the implications of quantum mechanics on particle behavior in collisions
  • Explore the derivation of the de Broglie wavelength and its applications
  • Study the principles of Rutherford backscattering in detail
  • Investigate the differences in particle behavior between alpha and beta decay
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Physicists, nuclear engineers, and students studying particle physics or quantum mechanics will benefit from this discussion, particularly those interested in the dynamics of alpha particles and their interactions with heavy atoms.

Orthoceras
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Maximum acceleration of an alpha particle that is backscattered by a heavy atom?
I would like to estimate the maximum acceleration (or deceleration) of an alpha particle that is backscattered by a heavy atom, like in Rutherford backscattering. I am interested in the order of magnitude, not in a precise value. I am assuming the collision is elastic.

The kinetic energy of the free alpha particle is E. The closest distance of the alpha particle to the atom's nucleus, r_0, is found by equating E to the electrostatic potential energy: E = \frac{f q Q}{r_0} \Rightarrow r_0 = \frac{f q Q}{E}
The acceleration (or deceleration) 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}

For example, if the alpha particle was generated by the decay of Po-210, with E=5.4 MeV, and the scattering nucleus is heavy compared to the alpha particle (for example, another polonium atom) then r_0 = 4⋅10^{-14} m and a_0 = 3⋅10^{27} \frac{m}{s^2}

However, I suppose a quantum mechanical wave does not have an acceleration, so r should be at least 10 times the de Broglie wavelength (assuming the de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle). I tried a few r values and found that r=10 \lambda_{Broglie} at r=1.7 r_0. Then the acceleration is a = 1⋅10^{27} \frac{m}{s^2}. That is my estimate for the order of magnitude of the maximum acceleration of this 5.4 MeV alpha particle.

Is this a somewhat valid estimate?
 
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Orthoceras said:
However, I suppose a quantum mechanical wave does not have an acceleration, so r should be at least 10 times the de Broglie wavelength (assuming the de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle). I tried a few r values and found that r=10 \lambda_{Broglie} at r=1.7 r_0. Then the acceleration is a = 1⋅10^{27} \frac{m}{s^2}. That is my estimate for the order of magnitude of the maximum acceleration of this 5.4 MeV alpha particle.

Is this a somewhat valid estimate?

An edit of that paragraph, to provide more clarity and correct a mistake:

However, the speed is zero at r=r_0, so the de Broglie wavelength is infinite, the alpha particle is a quantum mechanical wave, and I suppose a wave does not have an acceleration. Assuming the de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle, r should be at least 10 times the de Broglie wavelength. The de Broglie wavelength is \lambda=\frac{h}{p} and p=\sqrt{2 m (E-\frac{kqQ}{r})}. Then r\approx 10\, \lambda at r=2 r_0, and the acceleration is a = \left(\frac{r0}{r}\right)^2a_0 = 0.7⋅10^{27} \frac{m}{s^2}. That is my estimate for the order of magnitude of the maximum acceleration of this 5.4 MeV alpha particle.

Is this a somewhat valid estimate?
 
You are looking for an order of magnitude estimate. I don't understand why you are not content with the particle formulation of the answer. Why bother with the wave formulation in which you know that "maximum acceleration" is an ill-defined concept?
 
I bother with the wave formulation because of the next question. I was actually curious about the highest acceleration of objects present in our environment. The average human body contains 0.1 femtogram op Po-210, with an activity of 20 Bq, so many 5.4 MeV alpha particles are produced all the time inside us. The calculation of the maximum acceleration is the same as the one I used for Rutherford backscattering.

In this new case the wave is the initial form of the alpha particle, so it can't be ignored. My question is, at which distance do they transform from wave to particle, so at which distance is the concept of acceleration applicable? Would that be when r\approx 10\, \lambda, or at its first recoil action on the nucleus, or at its first interaction with an electron of the atom?

Even though an exact "maximum acceleration" is an ill-defined concept, I would expect that it should somehow be possible to estimate the order of magnitude.
 
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Why are you considering alphas and not betas?
 
In alpha decay, the initial speed of the particle is known, it is zero. In beta decay, the initial speed is unknown. As a result I was able to calculate the initial acceleration of the particle in alpha decay, but not in beta decay..

Also, Rutherford has shown that an alpha particle is able to penetrate the atom as a projectile. An electron that penetrates an atom is less simple.
 
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