Question

The Large Hadron Collider (LHC). Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Collider (LHC) at the CERN Laboratory in Geneva, Switzerland. The machine accelerates protons to high kinetic energies in an underground ring 27km in circumference. (a) What is the speed v of a proton in the LHC if the proton’s kinetic energy is 7.0 TeV ? (Because v is very close to c, write $\mathit{v}\mathbf{=}\left(1-∆\right)\mathit{c}$and give your answer in terms of $\mathbf{∆}$.) (b) Find the relativistic mass, ${\mathit{m}}_{\mathbf{r}\mathbf{e}\mathbf{l}}$ of the accelerated proton in terms of its rest mass.

Short answer

1. The speed of photon V in terms of $∆$is $9.053\times {10}^{-9}$.
2. The relativistic mass, $m_{rel}$of the accelerated proton in terms of its rest mass is $7451.39m_{rest}$.

Step-by-step solution

Define the kinetic energy and the rest energy

Kinetic energy is the difference between total energy and rest energy.

The total energy E of a particle is

$$\begin{gathered} E=K+m{c}^2 \\ =\frac{m{c}^2}{\sqrt{1-v^2/c^2}} \\ =\gamma m{c}^2 \end{gathered}$$

Where, K is the relativistic kinetic energy. The expression of K is:

$K=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}-m{c}^2$

$\gamma$ is the Lorentz factor. The expression of $\gamma$ is:

$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$

The energy $\mathit{m}{\mathit{c}}^{\mathbf{2}}$ associated with rest mass m rather than motion is known as rest energy of the particle.

Determine the speed.

The mass of proton is $m=1.67\times {10}^{-27}kg$.

Given that, $K=7.0TeV$.

The kinetic energy in J :

role="math" localid="1664104793297" $$\begin{gathered} K=\left(7\times {10}^{12}\right)\left(1.6\times {10}^{-19}\right) \\ =1.12\times {10}^{-6}J \end{gathered}$$

The speed of photon v is:

$$\begin{gathered} v=\sqrt[C]{1-{\left(\frac{m{c}^2}{K+mc}\right)}^2} \\ =\sqrt[C]{1-{\left(\frac{\left(1.6276\times {10}^{-27}\right)\left(3\times {10}^8\right)}{\left(1.12\times {10}^{-6}\right)+\left(1.6726\times {10}^{-27}\right)\left(3\times {10}^8\right)}\right)}^2} \\ =0.99c \end{gathered}$$

Also, given that $v=\left(1-∆\right)c$

$$\begin{gathered} ∆=\frac{c-v}{c} \\ =\frac{c-0.99c}{c} \\ =9.053\times {10}^{-9} \end{gathered}$$

Hence, the speed of photon v in terms of $∆$ is $9.053\times {10}^{-9}$.

Determine the relativistic mass mrel .

$$\begin{gathered} \sqrt{1-v^2/c^2}=\frac{m{c}^2}{K+m{c}^2} \\ =\frac{\left(1.6726\times {10}^{-27}\right)\left(3\times {10}^8\right)}{\left(1.12\times {10}^{-6}\right)+\left(1.6726\times {10}^{-27}\right)\left(3\times {10}^8\right)} \\ =1.342\times {10}^{-4} \end{gathered}$$

The relativistic mass $m_{rel}$is:

$$\begin{gathered} m_{rel}=\frac{m_{rest}}{\sqrt{1-v^2/c^2}} \\ =\frac{m_{rest}}{1.342\times {10}^{-4}} \\ =7451.39m_{rest} \end{gathered}$$

Hence, the relativistic mass, $m_{rest}$of the accelerated proton in terms of its rest mass is $7451.39m_{rest}$.