Question

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What is the kinetic energy of the antiproton? Note: m.c2 = 938.3 MeV.

Answer 1

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

$E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}$

We know that,

$m_{p}c^2=m_{np}c^2$

So, $E_{photon}=2mc^2+K.E_{p}+K.E_{np}$

$K.E_{np}=E_{photon}-(2mc^2+K.E_{p})$

Put the value into the formula

$K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}$

$K.E_{np}=147.4\ MeV$

Hence, The kinetic energy of the anti proton is 147.4 MeV.