Chapter 14, Problem 9P

(a)

To determine

Prove that the nonrelativistic particles obey the relation $K_{th}=-Q\left(1+\frac{M_a}{M_X}\right)$.

Answer to Problem 9P

Equation $K_{th}=-Q\left(1+\frac{M_a}{M_X}\right)$ is proved.

Explanation of Solution

Consider the CM frame. Draw the diagram showing direction of $a$ and $X$.

Write the equation for kinetic energy of system in CM frame.

    $\begin{array}{c}{K}_{\text{CM}}=\frac{p^2}{2M_a}+\frac{p^2}{2M_{\chi }}\\ =\frac{p^2}{2}\left[\frac{M_{\chi }+M_a}{M_a{M}_{\chi }}\right]\end{array}$        (I)

Here, $K_{\text{CM}}$ is the kinetic energy, $M_a$ is the mass of particle A, $M_{\chi }$ is the mass of particle X, and $p$ is the momentum.

Consider the reaction in lab frame. Draw the diagram showing direction of $a$ and $X$ in lab frame.

Consider the equation for momentum in lab frame.

    $\begin{array}{c}{P}_{\text{lab}}=M_a\left(v+V\right)\\ =p\left(\frac{M_{\chi }+M_a}{M_{\chi }}\right)\end{array}$

Write the equation for momentum in lab frame.

  $K_{\text{lab}}=\frac{p_{\text{lab}}^2}{2M_a}$

Rewrite the above equation by substituting $p\left(\frac{M_{\chi }+M_a}{M_{\chi }}\right)$ for $P_{\text{lab}}$.

    $K_{\text{lab}}=\frac{{\left(p\left(\frac{M_{\chi }+M_a}{M_{\chi }}\right)\right)}^2}{2M_A}$        (II)

Divide equation (II) by equation (I).

    $\begin{array}{c}\frac{K_{\text{lab}}}{K_{\text{CM}}}=\frac{\frac{{\left(p\left(\frac{M_{\chi }+M_a}{M_{\chi }}\right)\right)}^2}{2M_A}}{\left(\frac{p^2}{2}\left[\frac{M_{\chi }+M_a}{M_a{M}_{\chi }}\right]\right)}\\ K_{\text{lab}}=K_{\text{CM}}\left[\frac{M_{\chi }+M_a}{M_{\chi }}\right]\end{array}$

Write the expression for minimum kinetic energy from the calculation.

    $K_{th}=-Q\left(1+\frac{M_a}{M_X}\right)$

Conclusion:

Thus, the equation $K_{th}=-Q\left(1+\frac{M_a}{M_X}\right)$ is proved.

(b)

To determine

Obtain the threshold energy for the alpha particle for the given reaction.

Answer to Problem 9P

Minimum energy required will be $1.53\text{MeV}$.

Explanation of Solution

Write the equation to find the Q value.

  $Q=\left[m\left(\text{N}\right)+m\left(\text{H}\text{e}\right)-m\left(\text{O}\right)-m\left(\text{H}\right)\right]\left(931.5\text{MeV}/\text{u}\right)$

Here, $m\left(\text{N}\right)$ is the mass of $\text{N}$, $m\left(\text{H}\text{e}\right)$ is the mass of $\text{H}\text{e}$, $\text{O}$ is the mass of $m\left(\text{O}\right)$ and $m\left(\text{H}\right)$ is the mass of $\text{H}$.

Rewrite equation (III) by replacing $M_a$ with $m\left(\text{H}\text{e}\right)$ and $M_X$ with $m\left(\text{N}\right)$.

    $K_{th}=-Q\left(1+\frac{m\left(\text{H}\text{e}\right)}{m\left(\text{N}\right)}\right)$        (III)

Conclusion:

Substitute $14.003074\text{ u}$ for $m\left(\text{N}\right)$, $4.002603\text{ u}$ for $m\left(\text{H}\text{e}\right)$, $16.999132\text{ u}$ for $m\left(\text{O}\right)$ and $1.007825\text{ u}$ for $m\left(\text{H}\right)$ in the above equation.

  $\begin{array}{c}Q=\left[14.003074\text{ u}+4.002603\text{ u}-16.999132\text{ u}-1.007825\text{ u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =-1.19\text{ MeV}\end{array}$

Substitute $-1.19\text{ MeV}$ for $Q$, $14.003074\text{ u}$ for $m\left(\text{N}\right)$ and $4.002603\text{ u}$ for $m\left(\text{H}\text{e}\right)$ in equation (IV).

    $\begin{array}{c}{K}_{th}=-\left(-1.19\text{ MeV}\right)\left(1+\frac{4.002603\text{ u}}{4.002603\text{ u}}\right)\\ =1.53\text{MeV}\end{array}$

Thus, the minimum energy required will be $1.53\text{MeV}$.