91Ó°ÊÓ

In case of the annihilation if the mass gets converted into the energy. If we get value of total mass destroyed then the expression for the energy is obtained by,

\(E = m{c^2}\)

Here\(E\)is the energy,\(m\)is the mass and\(c\)is the speed of the light.

We are using the conservation of energy and then evaluate:

\(E = {m_p}{c^2}\)

Here\({m_p}\)is the mass of the proton,

Substitute\(1.67 \times {10^{ - 27}}\,{\rm{kg}}\)for\({m_p}\)and\(3 \times {10^8}\,{\rm{m/s}}\)for\(c\)in the above equation.

\(\begin{array}{c}E = \left( {1.67 \times {{10}^{ - 27}}} \right){\left( {3 \times {{10}^8}} \right)^2}\\ = 15.0536 \times {10^{ - 11}}\,{\rm{J}}\end{array}\)

Convert into\({\rm{MeV}}\)

\(\begin{array}{c}E = \left( {15.05 \times {{10}^{ - 11}}\,{\rm{J}}} \right)\left( {6.242 \times {{10}^{12}}\,\frac{{{\rm{MeV}}}}{{\rm{J}}}} \right)\\ = 939\,{\rm{MeV}}\end{array}\)

Therefore the characteristics of \(\gamma \) rays are found to be \(939\,{\rm{MeV}}\).

Now compare the above calculated value with \(0.511\,\,{\rm{MeV}}\),

\(\begin{array}{c}Ratio = \frac{{939}}{{0.511}}\\ = 1.84 \times {10^3}\end{array}\)

Therefore the proton-antiproton annihilation energy is \(1.84 \times {10^3}\) times greater than electron-positron annihilation.