Chapter 7, Problem 103P

(a)

To determine

The ratio of the speeds $v_{\alpha }/v_{\text{Rn}}$.

Answer to Problem 103P

The ratio of the speeds $v_{\alpha }/v_{\text{Rn}}$ is $55.5$,

Explanation of Solution

Write the expression for conservation of momentum.

  $p_i=p_f$        (I)

Here, $p_i$ is the initial momentum and $p_f$ is the final momentum.

Since the initial momentum of the system is zero

  $p_i=0$        (II)

The final momentum of the system is,

  $p_f=m_{Rn}{v}_{Rn}+m_{\alpha }{v}_{\alpha }$        (III)

Here, $m_{Rn}$ is the mass of the radon nucleus, $v_{Rn}$ is the speed of the radon nucleus decays, $m_{\alpha }$ is the mass of the alpha particle, and $v_{\alpha }$ is the speed of the alpha particle decays.

Conclusion:

Substitute the equation (II) and (III) in equation (I).

  $\begin{array}{c}{m}_{Rn}{v}_{Rn}+m_{\alpha }{v}_{\alpha }=0\\ m_{Rn}{v}_{Rn}=-m_{\alpha }{v}_{\alpha }\\ \frac{v_{\alpha }}{v_{Rn}}=-\frac{m_{Rn}}{m_{\alpha }}\\ \frac{\left|v_{\alpha }\right|}{\left|v_{Rn}\right|}=\frac{m_{Rn}}{m_{\alpha }}\end{array}$

Substitute $222\text{u}$ for $m_{Rn}$ and $4\text{u}$ for $m_{\alpha }$ in the above relation to find ratio $v_{\alpha }/v_{\text{Rn}}$.

  $\begin{array}{c}\frac{\left|v_{\alpha }\right|}{\left|v_{Rn}\right|}=\frac{\left(222\text{u}\right)}{4\text{u}}\\ =55.5\end{array}$

Therefore, the ratio of the speeds $v_{\alpha }/v_{\text{Rn}}$ is $55.5$,

(b)

To determine

The ratio of the magnitude of the momentum $p_{\alpha }/p_{\text{Rn}}$.

Answer to Problem 103P

The ratio of the magnitude of the momentum $p_{\alpha }/p_{\text{Rn}}$ is $1.0$.

Explanation of Solution

Write the expression for momentum of the alpha particle.

  $p_{\alpha }=m_{\alpha }{v}_{\alpha }$        (IV)

Here, $m_{\alpha }$ is the mass of the alpha particle and $v_{\alpha }$ is the speed of the alpha particle decays.

Write the expression for momentum of the radon particle.

  $p_{Rn}=m_{Rn}{v}_{Rn}$        (V)

Here, $m_{Rn}$ is the mass of the radon nucleus and $v_{Rn}$ is the speed of the radon nucleus decays,

Conclusion:

Divide the equation (IV) by (V).

  $\frac{p_{\alpha }}{p_{Rn}}=\frac{m_{\alpha }{v}_{\alpha }}{m_{Rn}{v}_{Rn}}$

Substitute $222\text{u}$ for $m_{Rn}$, $55.5$ for $v_{\alpha }/v_{\text{Rn}}$, and $4\text{u}$ for $m_{\alpha }$ in the above relation to find ratio $p_{\alpha }/p_{\text{Rn}}$.

  $\begin{array}{c}\frac{p_{\alpha }}{p_{Rn}}=\frac{\left(4\text{u}\right)}{\left(222\text{u}\right)}\left(55.5\right)\\ =1.0\end{array}$

Therefore, the ratio of the magnitude of the momentum $p_{\alpha }/p_{\text{Rn}}$ is $1.0$.

(c)

To determine

The ratio of the kinetic energies $K_{\alpha }/K_{\text{Rn}}$.

Answer to Problem 103P

The ratio of the kinetic energies $K_{\alpha }/K_{\text{Rn}}$ is $55.5$.

Explanation of Solution

Write the expression for kinetic energy of the alpha particle.

  $K_{\alpha }=\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2$        (VI)

Write the expression for kinetic energy of the radon particle.

  $K_{Rn}=\frac{1}{2}{m}_{Rn}{v}_{Rn}^2$        (VII)

Conclusion:

Divide the equation (VI) by (VII).

  $\begin{array}{c}\frac{K_{\alpha }}{K_{Rn}}=\frac{\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2}{\frac{1}{2}{m}_{Rn}{v}_{Rn}^2}\\ =\frac{m_{\alpha }}{m_{Rn}}{\left(\frac{v_{\alpha }}{v_{Rn}}\right)}^2\end{array}$

Substitute $222\text{u}$ for $m_{Rn}$, $55.5$ for $v_{\alpha }/v_{\text{Rn}}$, and $4\text{u}$ for $m_{\alpha }$ in the above relation to find ratio $K_{\alpha }/K_{\text{Rn}}$.

  $\begin{array}{c}\frac{K_{\alpha }}{K_{Rn}}=\frac{\left(4\text{u}\right)}{\left(222\text{u}\right)}{\left(55.5\right)}^2\\ =55.5\end{array}$

Therefore, the ratio of the kinetic energies $K_{\alpha }/K_{\text{Rn}}$ is $55.5$.