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Q15 PE

Expert-verified
Found in: Page 1149

College Physics (Urone)

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.

See the step by step solution

Step by Step Solution

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.

Step 2: Evaluating the velocity of beta particles

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}

Step 3: Evaluating the velocity of alpha particles

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}

Step 4: Evaluating the ratio

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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