Chapter 30, Problem 65GP

(a)

To determine

To show: The nucleus carries away a fraction $\frac{1}{\left(1+\frac{1}{4}{A}_D\right)}$ of the total energy available.

Answer to Problem 65GP

The nucleus carries away a fraction $\frac{1}{\left(1+\frac{1}{4}{A}_D\right)}$ of the total energy available.

Explanation of Solution

Given:

Alpha decay of a $\text{R}\text{a}$ nucleus

Formula used:

Kinetic energy can be given as:

  $KE=\frac{p^2}{2m}$

Calculation:

Now, using the law of conservation of momentum, it can be concluded that the momentum of the daughter nucleus will be equal to the momentum of alpha particle as follows:

  $p_{\alpha }=p_D\mathrm{...........................}(1)$

So, the kinetic energy of the daughter nucleus will be given as:

  $K{E}_D=\frac{p_D^2}{2m_D}$

Using equation (1), we get:

  $K{E}_D=\frac{p_{\alpha }^2}{2m_D}$

Now, the momentum of the alpha particle can be written as:

  $p_{\alpha }=\sqrt{2m_{\alpha }K{E}_{\alpha }}$

So, the previous equation becomes:

  $\begin{array}{l}K{E}_D=\frac{{\left(\sqrt{2m_{\alpha }K{E}_{\alpha }}\right)}^2}{2m_D}\\ K{E}_D=\frac{2m_{\alpha }K{E}_{\alpha }}{2m_D}\\ K{E}_D=\left(\frac{m_{\alpha }}{m_D}\right)K{E}_{\alpha }\\ K{E}_D=\left(\frac{A_{\alpha }}{A_D}\right)KE\end{array}$

Here, $A_{\alpha }$ is the mass number of alpha particle which is equal to 4.

  $A_D$ is the mass number of daughter nucleus.

  $\begin{array}{l}K{E}_D=\left(\frac{4}{A_D}\right)K{E}_{\alpha }\\ \frac{K{E}_D}{K{E}_{\alpha }+K{E}_D}=\frac{\left(\frac{4}{A_D}\right)}{\left(1+\frac{4}{A_D}\right)}\\ \frac{K{E}_D}{K{E}_{\alpha }+K{E}_D}=\frac{1}{\left(1+\frac{1}{4}{A}_D\right)}\end{array}$

Conclusion:

Hence, the ratio is equal to $\frac{1}{\left(1+\frac{1}{4}{A}_D\right)}$

(b)

To determine

To find: The percentage of the energy available.

Answer to Problem 65GP

The percentage carried away by the alpha particle is given by is 98.23%

Explanation of Solution

Given:

Alpha decay of a $\text{R}\text{a}$ nucleus

Calculation:

The decay equation can be written as:

  $\text{R}\text{a}\to \text{R}\text{n}+\text{H}\text{e}$

So, the mass number of the daughter nucleus is $A_D=222$

Using $\frac{K{E}_D}{K{E}_{\alpha }+K{E}_D}=\frac{1}{\left(1+\frac{1}{4}{A}_D\right)}$ from part (a) and putting the value $A_D=222$ , we get:

  $\frac{K{E}_D}{K{E}_{\alpha }+K{E}_D}=\frac{1}{\left(1+\frac{222}{4}\right)}=0.0177$

Hence, the percentage carried away by the alpha particle is given by:

  $\begin{array}{l}=1-0.0177\\ =0.9823\\ =98.23\%\end{array}$

Conclusion:

Hence, the percentage carried away by the alpha particle is given by is 98.23%