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

Answer:

K = 80.75 MeV    

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:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$  

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

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

$K_{ap} = 80.75 MeV$              

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

I hope it helps you!