1 Answer

Kusalananda · Answer 1 · 2020-10-29T14:26:14+0000

(a) 0.99999687c

The relationship between total energy (E) and rest energy (
E_0 ) is

$E=\gamma E_0$

where
$\gamma$ is the relativistic factor. In this problem, we know that the total energy is 400 times the rest energy: this means that

$\gamma=400$

The formula for the relativistic factor is

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

where v is the speed of the proton and c is the speed of light. By isolating the term v/c, we find the speed of the protons in terms of c:

$1-((v)/(c))^2=(1)/(\gamma^2)\\(v)/(c)=\sqrt{1-(1)/(\gamma^2)}=\sqrt{1-(1)/(400^2)}=0.99999687$

(b)
$3.75\cdot 10^5 MeV$

Given the rest mass of a proton:

$m_0=1.67\cdot 10^(-27)kg$

its rest energy is the energy equivalent to this mass:

$E_0=m_0 c^2 = (1.67\cdot 10^(-27)kg)(3\cdot 10^8 m/s)^2=1.5\cdot 10^(-10) J$

The protons in the Fermi Laboratory have energy that is 400 times their rest energy, so their total energy is

E=400 E_0

The total energy is also sum of rest energy and kinetic energy:

E=E_0+K

So the kinetic energy is

$K=E-E_0=400E_0 - E_0 = 399 E_0 = 399(1.5\cdot 10^(-10)J)=6.0\cdot 10^(-8) J$

And since the conversion factor is

$1 eV = 1.6\cdot 10^(-19) J$

This kinetic energy converted into MeV (
10^6 eV ) is

$K=(6.0\cdot 10^(-8)J)/(1.6\cdot 10^(-19)J/eV \cdot 10^6 eV/MeV)=3.75\cdot 10^5 MeV$

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