Question

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic energy of the proton is 242.85 MeV? Use the following Joules-to-electron-Volts conversion 1eV = 1.602 × 10-19 J. The rest mass of a proton is 1.67 × 10 − 27 kg.

Answer 1

K = 80.75 MeV

Step-by-step explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

`E_(ph) = E_(p) + E_(ap) = E_(0p) + K_(p) + E_(0ap) + K_(ap) = m_(0p)c^(2) + K_(p) + m_(0ap)c^(2) + K_(ap)`

where
`E_(ph)`: is the photon energy,
`E_(0p) and E_(0ap)`: are the rest energies of the proton and the antiproton, respectively, equals to m₀c²,
`K_(p) and K_(ap)`: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

`K_(ap) = E_(ph) - m_(0p)c^(2) - K_(p) - m_(0ap)c^(2)`

The proton mass is equal to the antiproton mass, so:

`K_(ap) = E_(ph) - 2m_(0p)c^(2) - K_(p)`

`K_(ap) = 2.20 \cdot 10^(3)MeV - 2(1.67 \cdot 10^(-27)kg)(3\cdot 10^(8) \frac {m}{s})^(2) - 242.85MeV`

`K_(ap) = 2.20 \cdot 10^(3)MeV - 2(1.67 \cdot 10^(-27)kg)(3\cdot 10^(8) \frac {m}{s})^(2)((1eV)/(1.602 \cdot 10^(-19)J))((1 MeV)/(10^(6)eV)) - 242.85MeV`

`K_(ap) = 80.75 MeV`

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!