(a) A particle and its antiparticle are at rest relative to an observer and annihilate (completely destroying both masses), creating two\({\rm{\gamma }}\)rays of equal energy. What is the characteristic\({\rm{\gamma }}\)-ray energy you would look for if searching for evidence of proton-antiproton annihilation? (The fact that such radiation is rarely observed is evidence that there is very little antimatter in the universe.) (b) How does this compare with the\({\rm{0}}{\rm{.511 MeV}}\)energy associated with electron-positron annihilation?

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.