Answer:
The kinetic energy of the anti proton is 147.4 MeV.
Explanation:
Given that,
Energy = 2.12 GeV
Kinetic energy = 96.0 MeV
We need to calculate the kinetic energy of the anti proton
Using formula of energy
[tex]E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}[/tex]
We know that,
[tex]m_{p}c^2=m_{np}c^2[/tex]
So, [tex]E_{photon}=2mc^2+K.E_{p}+K.E_{np}[/tex]
[tex]K.E_{np}=E_{photon}-(2mc^2+K.E_{p})[/tex]
Put the value into the formula
[tex]K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}[/tex]
[tex]K.E_{np}=147.4\ MeV[/tex]
Hence, The kinetic energy of the anti proton is 147.4 MeV.
