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Q11 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 $${\rm{\beta }}$$ 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: $$\frac{{{v_{beta}}}}{{{v_{alpha}}}} = 85.163$$.

See the step by step solution

## Step 1: Define Radioactivity

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 alpha particle

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:

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

## Step 3: Evaluating the beta particle

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}

## Step 3: Evaluating the ratio

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: $$85.163$$.