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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?
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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