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Q11 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 **\({\rm{\beta }}\)** particle to that of an**\({\rm{\alpha }}\)** particle, if they have the same nonrelativistic kinetic energy?**

The ratio of the velocity of both the particles is obtained as: \(\frac{{{v_{beta}}}}{{{v_{alpha}}}} = 85.163\).

**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:

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

Solve for the \({\rm{\alpha }}\) particle:

In order to evaluate the kinetic energy, we use the relation as:

\(K.{E_\alpha } = \frac{1}{2}{m_\alpha }v_{alpha}^2\)

Solve for the \({\rm{\beta }}\) particle:

In order to evaluate the kinetic energy, we use the relation as:

\(K.{E_\beta } = \frac{1}{2}{m_\beta }v_{beta}^2\)

As, the kinetic energy for both particles are equal, then we obtain:

\(\begin{align}K.{E_\alpha } &= K.{E_\beta }\\ &= \frac{1}{2}{m_\alpha }v_{alpha}^2\\ &= \frac{1}{2}{m_\beta }v_{beta}^2\end{align}\)

Rearranging and then solving to obtain the ratio of both the particles as:

\(\begin{align}\frac{{{v_{beta}}}}{{{v_{alpha}}}} &= \sqrt {\frac{{{m_{beta}}}}{{{m_{alpha}}}}} \\ &= \sqrt {\frac{{6.6 \times {{10}^{ - 27}}\,kg}}{{9.1 \times {{10}^{ - 31}}\,kg}}} \\ &= 85.163\end{align}\)

Therefore, the ratio is: \(85.163\).

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