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

The kinetic energy of the anti proton is 147.4 MeV.

Step-by-step 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*10^(9)-2*938.3*10^(6)-96*10^(6)`

`K.E_(np)=147.4\ MeV`

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