Chapter 31, Problem 60P

(a)

To determine

The minimum photon energy required to produce reaction in reference frame.

Answer to Problem 60P

The minimum photon energy required to produce reaction in reference frame is $E_{\gamma }=127\text{MeV}$.

Explanation of Solution

At threshold point, photon and a proton colliding head on to produce a proton and a pion at rest.

  $\text{p}+\gamma \to \text{p}+{\pi }^0$

Write the expression for the conservation of energy of the following reaction.

  $\frac{m_{\text{p}}{c}^2}{\sqrt{1-u^2/c^2}}+E_{\gamma }=m_{\text{p}}{c}^2+m_{\pi }{c}^2$        (I)

Here, $m_{\text{p}}$ is the mass of the proton, $m_{\pi }$ is the mass of the pion, $E_{\gamma }$ is the energy of the gamma ray, $c$ is the velocity of light, and $u$ is the velocity of the particle.

Write the expression for conservation of momentum.

  $\frac{m_{\text{p}}u}{\sqrt{1-u^2/c^2}}-\frac{E_{\gamma }}{c}=0$        (II)

Rewrite the above equation for $E_{\gamma }$.

  $E_{\gamma }=\frac{m_{\text{p}}{c}^2}{\sqrt{1-u^2/c^2}}\frac{u}{c}$        (III)

The energy of the each particle is,

  $\begin{array}{c}{m}_p{c}^2=938.3\text{MeV}\\ m_{\pi }{c}^2=135.0\text{MeV}\end{array}$

Conclusion:

Combining the equations (I) and (III).

  $\frac{m_{\text{p}}{c}^2}{\sqrt{1-u^2/c^2}}+\frac{m_{\text{p}}{c}^2}{\sqrt{1-u^2/c^2}}\frac{u}{c}=m_{\text{p}}{c}^2+m_{\pi }{c}^2$

Substitute $938.3\text{MeV}$ for $m_p{c}^2$ and $135.0\text{MeV}$ for $m_{\pi }{c}^2$ in the above relation to find $u/c$.

  $\begin{array}{c}\frac{938.3\text{MeV}}{\sqrt{1-u^2/c^2}}+\frac{938.3\text{MeV}}{\sqrt{1-u^2/c^2}}\frac{u}{c}=938.3\text{MeV}+135.0\text{MeV}\\ \frac{\left(938.3\text{MeV}\right)\left(1+u/c\right)}{\sqrt{\left(1-u/c\right)\left(1+u/c\right)}}=1073.3\text{MeV}\\ \frac{u}{c}=0.134\end{array}$

Substitute $0.134$ for $u/c$ and $938.3\text{MeV}$ for $m_p{c}^2$ in the equation (III) to find $E_{\gamma }$.

  $\begin{array}{c}{E}_{\gamma }=\frac{938.3\text{MeV}}{\sqrt{1-{\left(0.134\right)}^2}}\left(0.134\right)\\ =127\text{MeV}\end{array}$

Therefore, the minimum photon energy required to produce reaction in reference frame is $E_{\gamma }=127\text{MeV}$.

(b)

To determine

The wavelength of the photon.

Answer to Problem 60P

The wavelength of the photon is $1.06\text{mm}$.

Explanation of Solution

Write the expression from Wien’s displacement law.

  ${\lambda }_{\max }T=2.898\text{mm}\cdot \text{K}$

Here, ${\lambda }_{\max }$ is the maximum wavelength of the photon and $T$ is the absolute temperature.

Rewrite the above expression for ${\lambda }_{\max }$.

  ${\lambda }_{\max }=\frac{2.898\text{mm}\cdot \text{K}}{T}$        (IV)

Conclusion:

Substitute $2.73\text{K}$ for $T$ in the equation (IV) to find ${\lambda }_{\max }$.

  $\begin{array}{c}{\lambda }_{\max }=\frac{2.898\text{mm}\cdot \text{K}}{2.73\text{K}}\\ =1.06\text{mm}\end{array}$

Therefore, the wavelength of the photon is $1.06\text{mm}$.

(c)

To determine

The energy of the photon.

Answer to Problem 60P

The energy of the photon is $1.17\times {10}^{-3}\text{eV}$.

Explanation of Solution

Write the expression for energy of the photon.

  $E=\frac{hc}{\lambda }$        (V)

Here, $h$ is the plank’s constant, $c$ is the velocity of light, and $E$ is the energy of the photon.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $1.06\text{mm}$ for $\lambda$ in the equation (V) to find $E$

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(1.06\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(1.06\times {10}^{-3}\text{m}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.06\times {10}^{-3}\text{m}}\\ =1.17\times {10}^{-3}\text{eV}\end{array}$

Therefore, the energy of the photon is $1.17\times {10}^{-3}\text{eV}$.

(d)

To determine

The energy of the proton in the reference frame mentioned in part (a).

Answer to Problem 60P

The energy of the proton in the reference frame mentioned in part (a) is $5.81\times {10}^{19}\text{eV}$.

Explanation of Solution

Write the expression from Doppler effect.

  $E^{\prime }=\sqrt{\frac{1+v/c}{1-v/c}}E$        (VI)

Here, $E^{\prime }$ is the energy of the photon moving to left, $E$ is the energy of the photon in unprimed frame, and $v$ is the velocity of the source.

The speed of the proton is,

  $\frac{u}{c}=\frac{u^{\prime }/c+v/c}{1+u^{\prime }v/c^2}$        (VII)

Here, $u^{\prime }$ is the velocity of the proton in reference frame.

The energy of the proton is,

  $E=\frac{m_{\text{p}}{c}^2}{\sqrt{1-u^2/c^2}}$        (VIII)

Conclusion:

Substitute $1.27\times {10}^8\text{eV}$ for $E^{\prime }$ and $1.17\times {10}^{-3}\text{eV}$ for $E$ in equation (VI) to find $v/c$.

$\begin{array}{c}\left(1.27\times {10}^8\text{eV}\right)=\sqrt{\frac{1+v/c}{1-v/c}}\left(1.17\times {10}^{-3}\text{eV}\right)\\ \frac{v}{c}=1-1.71\times {10}^{-22}\end{array}$

Substitute $0.134$ for $u^{\prime }/c$ and $1-1.71\times {10}^{-22}$ for $v/c$ in equation (VII) to find $u/c$.

  $\begin{array}{c}\frac{u}{c}=\frac{0.134+1-1.71\times {10}^{-22}}{1+0.134\left(1-1.71\times {10}^{-22}\right)}\\ =1-1.30\times {10}^{-22}\end{array}$

Substitute $938.3\text{MeV}$ for $m_p{c}^2$ and $1-1.30\times {10}^{-22}$ for $u/c$ in equation (VIII) to find $E$.

  $\begin{array}{c}E=\frac{\left(938.3\text{MeV}\right)}{\sqrt{1-{\left(1-1.30\times {10}^{-22}\right)}^2}}\\ =6.19\times {10}^{10}\left(938.3\text{MeV}\right)\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\\ =5.81\times {10}^{19}\text{eV}\end{array}$

Therefore, the energy of the proton in the reference frame mentioned in part (a) is $5.81\times {10}^{19}\text{eV}$