Chapter 26, Problem 93P

(a)

To determine

Speed of movement of Polonium-218 nucleus.

Answer to Problem 93P

Speed of particle is $2.98\times {10}^5\text{m}/\text{s}$.

Explanation of Solution

Write the equation for law of conservation of energy for the given system.

  $m_{Ra}{c}^2=m_{\alpha }{c}^2+m_{Po}{c}^2+K_{\alpha }+K_{Po}$                                                                             (I)

Here, $m_{Ra}$ is the mass of radon-222 nucleus, $c$ is the speed of light in vacuum, $m_{\alpha }$ is the mass of alpha particle, $m_{Po}$ is the mass of polonium-218 nucleus, $K_{\alpha }$ is the kinetic energy of alpha particle, and $K_{Po}$ is the kinetic energy of polonium-218 nucleus.

Write the equation to find $K_{\alpha }$.

  $K_{\alpha }=\frac{p^2}{2m_{\alpha }}$                                                                                                               (II)

Here, $p$ is the momentum.

Write the equation to find $K_{Po}$.

  $K_{Po}=\frac{p^2}{2m_{Po}}$                                                                                                           (III)

Rewrite equation (I) by subtitling equations (II) and (III).

  $\begin{array}{c}{m}_{Ra}{c}^2=m_{\alpha }{c}^2+m_{Po}{c}^2+\frac{p^2}{2m_{\alpha }}+\frac{p^2}{2m_{Po}}\\ \left(m_{Ra}-m_{\alpha }-m_{Po}\right)c^2=p^2\left(\frac{1}{2m_{\alpha }}+\frac{1}{2m_{Po}}\right)\end{array}$

Rewrite the above expression in terms of $p$.

    $p=\sqrt{\frac{\left(m_{Ra}-m_{\alpha }-m_{Po}\right)c^2}{\left(\frac{1}{2m_{\alpha }}+\frac{1}{2m_{Po}}\right)}}$                                                                                     (IV)

Write the equation to find the speed of Polonium-218 nucleus.

    $v_{Po}=\frac{p}{m_{Po}}$                                                                                                              (V)

Here, $v_{Po}$ is the speed of Polonium-218 nucleus.

Conclusion:

Substitute $221.97039u$ for $m_{Ra}$, $4.00151u$ for $m_{\alpha }$, and $217.96289u$ for $m_{Po}$ in equation (IV) to find $p$.

    $\begin{array}{c}p=\sqrt{\frac{\left(221.97039\text{u}-4.00151\text{u}-217.96289\text{u}\right)c^2}{\left(\frac{1}{2\left(4.00151\text{u}\right)}+\frac{1}{2\left(217.96289\text{u}\right)}\right)}}\\ =\sqrt{\frac{\left(0.00599\text{u}\right)}{0.1272}}\\ =\left(0.217\text{u}\right)c\end{array}$

Substitute $\left(0.217\text{u}\right)c$ for $p$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$, and $217.96289$ for $m_{Po}$ in equation (V) to find $v_{Po}$.

  $\begin{array}{c}{v}_{Po}=\frac{\left(0.217\text{u}\right)\left(2.998\times {10}^8\text{m}/\text{s}\right)}{217.96289\text{u}}\\ =2.98\times {10}^5\text{m}/\text{s}\end{array}$

Therefore, the speed of particle is $2.98\times {10}^5\text{m}/\text{s}$.

(b)

To determine

Speed of movement of alpha particle.

Answer to Problem 93P

Speed of particle is $1.63\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Write the equation to find the speed of alpha particle.

    $v_{\alpha }=\frac{p}{m_{\alpha }}$                                                                                                              (VI)

Here, $v_{Po}$ is the speed of alpha particle.

Conclusion:

Substitute $\left(0.217\text{u}\right)c$ for $p$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$, and $4.00151\text{u}$ for $m_{Po}$ in equation (V) to find $v_{Po}$.

  $\begin{array}{c}{v}_{Po}=\frac{\left(0.217\text{u}\right)\left(2.998\times {10}^8\text{m}/\text{s}\right)}{217.96289\text{u}}\\ =2.98\times {10}^5\text{m}/\text{s}\end{array}$

Therefore, the speed of particle is $1.63\times {10}^7\text{m}/\text{s}$.