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Chapter 11: Problem 73

A proton and an antiproton collide head-on, with each having a kinetic energy of \(7.00 \mathrm{TeV}\) (such as in the \(\mathrm{LHC}\) at CERN). How much collision energy is available, taking into account the annihilation of the two masses? (Note that this is not significantly greater than the extremely relativistic kinetic energy.)

Short Answer

Expert verified
The total collision energy available after a head-on collision between a proton and an antiproton, each having kinetic energy of \(7.00\, TeV\), is approximately \(2.27 \times 10^{-6} J\). This takes into account both the kinetic energy and the energy due to mass annihilation of the two particles.

Step by step solution

01

Find the mass of a proton and an antiproton

The proton and antiproton have the same mass, which is approximately \(1.67 \times 10^{-27} kg\). We will denote this mass as \(m\). \(m = 1.67 \times 10^{-27} kg\)
02

Find the energy due to mass annihilation of a proton and an antiproton

Using the mass-energy equivalence formula, we'll find the energy due to mass annihilation. The speed of light, c, is approximately \(3.00 \times 10^8 m/s\). \(E = mc^2\) \(E = (1.67 \times 10^{-27} kg)(3.00 \times 10^8 m/s)^2\) \(E = 1.50 \times 10^{-10} J\) Since there are two particles (proton and antiproton), the total energy due to mass annihilation will be twice this value: \(E_{total} = 2 \times 1.50 \times 10^{-10} J\) \(E_{total} = 3.00 \times 10^{-10} J\)
03

Convert the kinetic energy from TeV to Joules

The kinetic energy of each particle is given in TeV. To convert it to Joules, we'll use the conversion factor \(1 TeV = 1.60 \times 10^{-7} J\). \(7.00\, TeV = (7.00)(1.60 \times 10^{-7} J)\) \(7.00\, TeV = 1.12 \times 10^{-6} J\)
04

Find the total kinetic energy of the two particles

Since the proton and antiproton have the same kinetic energy, the total kinetic energy (\(K_{total}\)) will be the sum of their individual kinetic energies: \(K_{total} = (1.12 \times 10^{-6} J) + (1.12 \times 10^{-6} J)\) \(K_{total} = 2.24 \times 10^{-6} J\)
05

Calculate the total collision energy available

Now, we'll add the total energy due to mass annihilation (\(E_{total}\)) and the total kinetic energy (\(K_{total}\)) to find the total collision energy available (\(E_{collision}\)): \(E_{collision} = E_{total} + K_{total}\) \(E_{collision} = (3.00 \times 10^{-10} J) + (2.24 \times 10^{-6} J)\) \(E_{collision} = 2.27 \times 10^{-6} J\) The total collision energy available is approximately \(2.27 \times 10^{-6} J\).

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