Chapter 37: Q. 28 (page 1082)

The Large Hadron Collider accelerates two beams of protons, which travel around the collider in opposite directions, to a total energy of $6.5 TeV$ per proton. ( $1 TeV = 1$ teraelectron volt $10^{12} e V$ ) The beams cross at several points, and a few protons undergo headon collisions. Such collisions usually produce many subatomic particles, but in principle the colliding protons could produce a single subatomic particle at rest. (It would be unstable and would almost instantly decay into other subatomic particles.) What would be the mass, as a multiple of the proton's mass, of such a particle?

Short Answer

Expert verified

The produced particle would have the mass $14000$ times the mass of the proton.

Step by step solution

Given Information

The energy of a single proton is given to be $E_{p} = 6.5 TeV$

Calculation

If the energy of a single proton is $E_{p} = 6.5 TeV$ , then two head on colliding protons have the total energy of $E_{1} = 2 E_{p}$ . After the collislon a particle at rest is produced so it only has a rest energy $E_{2} = m c^{2}$ , where $m$ is the mass of that particle. Due to energy conservation, the energy before and after collision must be the same i.e. $E_{1} = E_{2}$ so we have
$2 E_{p} = m c^{2},$
yielding
$m = \frac{2 E_{p}}{c^{2}} = 13 TeV / c^{2}$
From Table $42.2$ in the book we read that the mass of the proton is $m_{p} = 938.28 MeV / c^{2}$ . Therefore
$\frac{m}{m_{p}} = \frac{13 \cdot 10^{12} eV / c^{2}}{938.28 \cdot 10^{6} eV / c^{2}} = 14000$
In other words
$m = 14000 m_{p}$

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