Question

Physicists and engineers from around the world have come 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 27 km in circumference.

a. What speed v of proton in the LHC if the proton's kinetic energy is 6.7 TeV? (Because v is very close to c, write v=(1−Δ)c and give your answer in terms of Δ).
b. Find the relativistic mass, mrel, of the accelerated protons in terms of their rest mass.

Answer 1

The speed of the proton in the LHC can be calculated using the relativistic energy equation, and the value is expressed as v = (1 - Δ)c. The relativistic mass of the proton can also be determined using the equation mrel = γm, where the Lorentz factor γ is calculated based on the kinetic energy of the proton.

Step-by-step explanation:

In the Large Hadron Collider (LHC), the protons are accelerated to high kinetic energies. To find the speed of the protons, we can use the relativistic energy equation: E = (γ - 1)mc^2 where γ is the Lorentz factor and m is the rest mass of the proton. Since the kinetic energy is given as 6.7 TeV, we can equate the two equations to find the value of γ.

Solving for γ, we get: γ = 1 + E/(mc^2) = 1 + (6.7 TeV)/(938 MeV).

Now, we can find the speed of the proton using the equation: v = c(1 - 1/γ).

Therefore, the speed of the proton v in the LHC is approximately v = (1 - Δ)c, where Δ is equal to 1/γ.

b. To find the relativistic mass of the proton, we can use the equation: mrel = γm, where m is the rest mass of the proton. Substituting the value of γ calculated earlier, we get:

mrel = (1 + (6.7 TeV)/(938 MeV)) * m.

Answer 2

Energy of photon, E = 6.7 eV

E =
`$6.7 * 1.602 * 10^(-7)$` joule

Kinetic energy,
`$K.E. =(1)/(2) mv^2 = 1.602 * 6.7 * 10^(-7)$`

`$v^2=(2 * 1.602 * 6.7 * 10^(-7))/(1.6726 * 10^(-27))$`

`$=12.834 * 10^(-20)$`

Kinetic energy at high speeds

`$(r-1)* mc^2 = 6.7 \ eV$`

`$(r-1)=(6.7 * 1.602 * 10^(-7))/(1.6726 * 10^(-27) * 9 * 10^(16))$`

r - 1 = 7130

r = 7130 + 1

r = 7131

`$\frac{1}{\sqrt{1-(v^2)/(C^2)}}=7131$`

`$1-(v^2)/(C^2) = \left((1)/(7131)\right)^2$`

`$v^2=C^2\left[1-\left((1)/(7131)\right)^2\right]$`

`$v=0.99999999017C$`

Δ = 1 - 0.99999999017

= 0.00000000933

Relative mass,
`$m_(rel)=r.m$`

`$=7131 * 1.6728 * 10^(-27)$`

`$=1.1927 * 10^(-23)$` kg