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Q15 PE

Expert-verifiedFound in: Page 1149

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**What is the ratio of the velocity of a \(5.00\,{\rm{MeV }}\beta \)ray to that of an\({\rm{\alpha }}\) particle with the same kinetic energy? This should confirm that \({\rm{\beta }}\) s travel much faster than \({\rm{\alpha }}\) s even when relativity is taken into consideration. (See also Exercise \({\rm{31}}{\rm{.11}}\).)**

The velocity of \({\rm{\beta }}\) particles obtained is nineteen times the velocity of \({\rm{\alpha }}\) particle.

**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.**

The mass of the \({\rm{\beta }}\) particle is equal to the value of:

\({m_\beta } = 9.1 \times {10^{ - 31}}\,{\rm{kg}}\)

The mass of the \({\rm{\alpha }}\) particle is equal to:

\({m_\alpha } = 6.6 \times {10^{ - 27}}\,{\rm{kg}}\)

As, the \({\rm{\beta }}\) particle is moving with high velocity.

We then solve for\({\rm{\beta }}\)particle.

To evaluate the kinetic energy, we use the relation as:

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

Rearranging and solving for the relativistic factor for \({\rm{\beta }}\) particle as:

\(\begin{align}{\gamma _\beta } &= \frac{{K \times {E_\beta }}}{{{m_\beta }{c^2}}} + 1\\ &= \frac{{5\,MeV}}{{0.511\,MeV/{c^2} \times {c^2}}} + 1\\ &= 10.785\end{align}\)

The relation used to evaluate the relativistic factor is:

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

Rearranging and solving the velocity of beta particles as:

\(\begin{align}{v_\beta }{\rm{ }} &= c\sqrt {1 - \frac{1}{{\gamma _\beta ^2}}} \\ &= 3 \times {10^8}\,m/s\sqrt {1 - \frac{1}{{{{(10.785)}^2}}}} \\ &= 2.987 \times {10^8}\,m/s\end{align}\)

We then solve for \({\rm{\alpha }}\)particle.

To evaluate the kinetic energy, we use the relation as:

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

Rearranging and solving for the relativistic factor for \({\rm{\alpha }}\) particle as:

\(\begin{align}{\gamma _\alpha }{\rm{ }} &= \frac{{K \times {E_\alpha }}}{{{m_\alpha }{c^2}}} + 1\\ &= \frac{{5\,MeV}}{{(4\,u \times \frac{{931.5\,MeV/u}}{{{c^2}}} \times {c^2}}} + 1\\ &= 1.00134\end{align}\)

The relation used to evaluate the relativistic factor is:

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

Rearranging and solving the velocity of alpha particles as:

\(\begin{align}{v_\alpha }{\rm{ }} &= c\sqrt {1 - \frac{1}{{\gamma _\alpha ^2}}} \\ &= 3 \times {10^8}\,m/s\sqrt {1 - \frac{1}{{{{(1.00134)}^2}}}} \\ &= 1.552 \times {10^7}\,m/s\end{align}\)

Rearranging and then solving for the ratio between the velocity of \({\rm{\beta }}\) particle and the velocity of \({\rm{\alpha }}\) particle as:

\(\begin{align}\frac{{{v_{beta}}}}{{{v_{alpha}}}} &= \frac{{2.987 \times {{10}^8}\,m/s}}{{1.552 \times {{10}^7}\,m/s}}\\ &= 19.246\end{align}\)

Therefore, the beta particle velocity found is nineteen times that of the alpha particle.

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