Question

(a) A high-energy beam of alpha particles collides with a stationary helium gas target. What must the total energy of a beam particle be if the available energy in the collision is $16.0GeV$? (b) If the alpha particles instead interact in a colliding-beam experiment, what must the energy of each beam be to produce the same available energy?

Short answer

The total energy of a beam particle is $E_m=30.707MeV$and the energy of each beam be to produce the same available energy is$E_m=8.00GeV$
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Step-by-step solution

Definition of Law of conservation of energy

It states that energy can neither be created nor be destroyed but converted into one form to another.

Determine the total energy of a beam particle and the energy of each beam be to produce the same available energy

The mass of the alpha particle is $m_{\alpha }=6.666\times {10}^{-27}kg$and the rest energy of an alpha particle is $m_{\alpha }{c}^2=3.718GeV$

When a stationary particle is bombarded with an accelerated particle with total energy $E_m$where both particles have equal masses, the available energy is given by

$E_m=2m{c}^2\left(E_m+m{c}^2\right)$

role="math" localid="1663922415949" $$\begin{gathered} {E}_m=\left(E_{\alpha }^2\right)/\left(2m{c}^2\right) \\ {E}_m=(16GeV{)}^2/2(3.718GeV)-3.718GeV \\ {E}_m=30.707GeV \end{gathered}$$

Hence, the total energy of a beam particle is$E_m=30.707MeV$

In order to produce same available energy each beam will carry half of the available energy $E_{\alpha }=8.00GeV$.