Question

The Large Hadron Collider accelerates two beams of protons, which travel around the collider in opposite directions, to a total energy of $6.5\mathrm{TeV}$per proton. ( $1\mathrm{TeV}=1$teraelectron volt ${10}^{12}eV$) 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

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 $E_p=6.5\mathrm{TeV}$

Calculation

If the energy of a single proton is $E_p=6.5\mathrm{TeV}$, then two head on colliding protons have the total energy of $E_1=2E_p$. 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
$2E_p=m{c}^2,$
yielding
$m=\frac{2E_p}{c^2}=13\mathrm{TeV}/c^2$
From Table $42.2$ in the book we read that the mass of the proton is $m_p=938.28\mathrm{MeV}/{\mathrm{c}}^2$. Therefore
$\frac{m}{m_p}=\frac{13·{10}^{12}\mathrm{eV}/{\mathrm{c}}^2}{938.28·{10}^6\mathrm{eV}/{\mathrm{c}}^2}=14000$
In other words
$m=14000m_p$