Chapter 31, Problem 30P

(a)

To determine

The expression for the threshold energy.

Answer to Problem 30P

The expression for the threshold energy is $\underset{\_}{K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}}$.

Explanation of Solution

Write the expression from the conservation of energy.

$E_{\min }+m_2{c}^2=\sqrt{{\left(m_3{c}^2\right)}^2+{\left(p_{3}c\right)}^2}$        (I)

Here, $E_{\min }$ is the energy, $c$ is the speed of light, $m_2,m_3$ are the masses and $p_3$ is the momentum.

The momentum of the fist particle is equal to the momentum of total mass.

Write the expression for the momentum.

$p_1=p_3$        (II)

Here, $p_1$ is the momentum.

Write the expression from the momentum conservation.

$\begin{array}{c}{\left(p_{1}c\right)}^2={\left(p_{3}c\right)}^2\\ =E_{\min }^2-{\left(m_1{c}^2\right)}^2\end{array}$        (III)

Here, $m_1$ is the mass.

Write the expression for the threshold energy.

$K_{\min }=E_{\min }-m_1{c}^2$        (IV)

Here, $K_{\min }$ is the threshold energy.

Conclusion:

Substitute, $\left(E_{\min }^2-{\left(m_1{c}^2\right)}^2\right)$ for ${\left(p_{3}c\right)}^2$ in equation (I) and squaring both side to find $E_{\min }$.

$\begin{array}{l}{E}_{\min }+m_2{c}^2=\sqrt{{\left(m_3{c}^2\right)}^2+\left(E_{\min }^2-{\left(m_1{c}^2\right)}^2\right)}\\ {\left[E_{\min }+m_2{c}^2\right]}^2={\left(m_3{c}^2\right)}^2+\left(E_{\min }^2-{\left(m_1{c}^2\right)}^2\right)\\ E_{\min }=\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2}{2m_2}\end{array}$

Substitute, $\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2}{2m_2}$ for $E_{\min }$ in equation (IV) to find $K_{\min }$.

$\begin{array}{c}{K}_{\min }=\left[\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2}{2m_2}\right]-m_1{c}^2\\ =\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}\end{array}$

Thus, the expression for the threshold energy is $\underset{\_}{K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}}$.

(b)

To determine

The value for threshold energy.

Answer to Problem 30P

The value for threshold energy is $\underset{\_}{5.63\text{GeV}}$.

Explanation of Solution

Write the expression for the threshold energy.

$K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$        (V)

Conclusion:

Substitute, $4\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_3$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_1$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_2$ in equation (V) to find $K_{\min }$.

$\begin{array}{c}{K}_{\min }=\frac{\left[{\left\{4\left(938.3{\text{MeV/c}}^{\text{2}}\right)\right\}}^2-{\left(\left(938.3{\text{MeV/c}}^{\text{2}}\right)+\left(938.3{\text{MeV/c}}^{\text{2}}\right)\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^{\text{2}}\right)}\\ =5.63\text{GeV}\end{array}$

Thus, The value for threshold energy is $\underset{\_}{5.63\text{GeV}}$.

(c)

To determine

The value for threshold energy.

Answer to Problem 30P

The value for threshold energy is $\underset{\_}{768\text{MeV}}$.

Explanation of Solution

Write the expression for the threshold energy.

$K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$        (V)

The value of the product for this reaction is,

$\begin{array}{c}{m}_3=\left(497.7{\text{MeV/c}}^{\text{2}}+1115.6{\text{MeV/c}}^{\text{2}}\right)\\ =1613.3{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Substitute, $\left(1613.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_3$, $\left(139.6{\text{MeV/c}}^{\text{2}}\right)$ for $m_1$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_2$ in equation (V) to find $K_{\min }$.

$\begin{array}{c}{K}_{\min }=\frac{\left[{\left\{\left(1613.3{\text{MeV/c}}^{\text{2}}\right)\right\}}^2-{\left(\left(139.6{\text{MeV/c}}^{\text{2}}\right)+\left(938.3{\text{MeV/c}}^{\text{2}}\right)\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^{\text{2}}\right)}\\ =786\text{MeV}\end{array}$

Thus, the value for threshold energy is $\underset{\_}{768\text{MeV}}$.

(d)

To determine

The value for threshold energy.

Answer to Problem 30P

The value for threshold energy is $\underset{\_}{280\text{MeV}}$.

Explanation of Solution

Write the expression for the threshold energy.

$K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$        (V)

The value of the product for this reaction is,

$\begin{array}{c}{m}_3=\left[2\left(938.3{\text{MeV/c}}^{\text{2}}\right)+135{\text{MeV/c}}^{\text{2}}\right]\\ =2011.6{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Substitute, $\left(2011.6{\text{MeV/c}}^{\text{2}}\right)$ for $m_3$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_1$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_2$ in equation (V) to find $K_{\min }$.

$\begin{array}{c}{K}_{\min }=\frac{\left[{\left\{\left(2011.6{\text{MeV/c}}^{\text{2}}\right)\right\}}^2-{\left(\left(938.3{\text{MeV/c}}^{\text{2}}\right)+\left(938.3{\text{MeV/c}}^{\text{2}}\right)\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^{\text{2}}\right)}\\ =280\text{MeV}\end{array}$

Thus, the value for threshold energy is $\underset{\_}{280\text{MeV}}$.

(e)

To determine

The value for threshold energy.

Answer to Problem 30P

The value for threshold energy is $\underset{\_}{4.43\text{TeV}}$.

Explanation of Solution

Write the expression for the threshold energy.

$K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$        (V)

Conclusion:

Substitute, $\left(91.2\times {10}^3{\text{MeV/c}}^{\text{2}}\right)$ for $m_3$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_1$, $\left(938.3{\text{MeV/c}}^{\text{2}}\right)$ for $m_2$ in equation (V) to find $K_{\min }$.

$\begin{array}{c}{K}_{\min }=\frac{\left[{\left\{\left(91.2\times {10}^3{\text{MeV/c}}^{\text{2}}\right)\right\}}^2-{\left(\left(938.3{\text{MeV/c}}^{\text{2}}\right)+\left(938.3{\text{MeV/c}}^{\text{2}}\right)\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^{\text{2}}\right)}\\ =4.43\text{TeV}\end{array}$

Thus, the value for threshold energy is $\underset{\_}{\text{4}\text{.43}\text{TeV}}$.