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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, [itex]r_0[/itex], is found by equating E to the electrostatic potential energy: [itex] E = \frac{f q Q}{r_0} \Rightarrow r_0 = \frac{f q Q}{E}[/itex]

The acceleration (or deceleration) at [itex]r = r_0[/itex], due to the repulsive Coulomb force, is: [itex]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} [/itex]

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 [itex]r_0 = 4⋅10^{-14} m[/itex] and [itex]a_0 = 3⋅10^{27} \frac{m}{s^2}[/itex]

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 [itex]r=10 \lambda_{Broglie}[/itex] at [itex]r=1.7 r_0[/itex]. Then the acceleration is [itex]a = 1⋅10^{27} \frac{m}{s^2}[/itex]. 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?

The kinetic energy of the free alpha particle is E. The closest distance of the alpha particle to the atom's nucleus, [itex]r_0[/itex], is found by equating E to the electrostatic potential energy: [itex] E = \frac{f q Q}{r_0} \Rightarrow r_0 = \frac{f q Q}{E}[/itex]

The acceleration (or deceleration) at [itex]r = r_0[/itex], due to the repulsive Coulomb force, is: [itex]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} [/itex]

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 [itex]r_0 = 4⋅10^{-14} m[/itex] and [itex]a_0 = 3⋅10^{27} \frac{m}{s^2}[/itex]

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 [itex]r=10 \lambda_{Broglie}[/itex] at [itex]r=1.7 r_0[/itex]. Then the acceleration is [itex]a = 1⋅10^{27} \frac{m}{s^2}[/itex]. 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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