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Find the maximum energy of alpha particles emitted by accelerated beam of Po-210

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The alpha decay of Po-210 to Pb-206 releases approximately 5.4 MeV of energy, of which 5.3 MeV is the kinetic energy of the alpha particle and 0.1 MeV is the recoil energy of Pb-206. Suppose that we have a beam of Po-210 ions with a kinetic energy of 100 TeV incident on a target. Occasionally a Po-210 ion will decay and emit an alpha particle.
What is the maximum kinetic energy of such an alpha particle in the rest frame of the target?

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Solution Summary

A detailed solution using Lorentz transforms for energy and momentum is given.

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The beam will consist of fully ionized Po-ions, the mass of 84 times ionized Po-210 is 195.55 GeV. These ions are accelerated such that they obtain a kinetic energy of 100 TeV. This means that the velocity of the ions in the beam, v, is such that the gamma factor is given by:

gamma_v = 1 + 100 TeV/195.55 GeV = 512.37

The speed is very close to c:

|v| = c sqrt(1-1/gamma^2) = approximately c[1 - 1/(2 gamma^2)] = c [1 - 1.9*10^(-6)]

The ...

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