91Ó°ÊÓ

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

\(K.{E_\beta } = \left( {{\gamma _\beta } - 1} \right){m_\beta }{c^2}\) …(1)

The value of \(K.{E_\beta }\) is kinetic energy.

The value of \({m_\beta }\) is mass of \({\rm{\beta }}\) particle.

The value of \({v_\beta }\) is the velocity of \({\rm{\beta }}\)particle.

The relation used to evaluate the relativistic factor is:

\({\gamma _\beta } = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)

Substitute all the value in the above equation.

\(\begin{align}{\gamma _\beta } &= \frac{1}{{\sqrt {1 - \frac{{{{\left( {0.998c} \right)}^2}}}{{{c^2}}}} }}\\{\gamma _\beta } &= 15.819\end{align}\)

Now putting the required value in the above equation and solving it to evaluate the value of kinetic energy as:

\(\begin{align}K.{E_\beta } &= \left( {15.819 - 1} \right) \times 9.1 \times {10^{ - 31}}\,kg \times {\left( {3 \times {{10}^8}\,{m \ mathord{\left/ {\vphantom {m s}} \right. \ } s}} \right)^2}\\ &= 1.2137 \times {10^{ - 12}}\,J \times \left( {\frac{{1\,MeV}}{{1.6 \times {{10}^{ - 13}}\,J}}} \right)\\ &= 7.585\,MeV\end{align}\)

Therefore, the kinetic energy of beta particle is:\(7.585\,{\rm{MeV}}\).

So, we see that:\({V_\gamma } = c\).

Then, the relative velocity is given by:

\(\begin{align}\frac{{Velocity{\rm{ }}of{\rm{ }}\gamma }}{{Velocity{\rm{ }}of{\rm{ }}\beta }} &= \frac{c}{{0.998c}}\\ &= 1.002\end{align}\)

Therefore, the relation is: \(1.002\).