Chapter 46, Problem 70CP

(a)

To determine

Find the minimum photon energy required.

Answer to Problem 70CP

The minimum photon energy required is $\text{127 MeV}$.

Explanation of Solution

Given decay reaction is,

    $p+\gamma \to p+{\pi }^0$

From energy conservation,

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

Here, $E_{\gamma }$ is the energy photon, $m_p$ is the mass of proton, $m_{\pi }$ is the mass of pion and $c$ is the speed of light.

From momentum conservation,

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

Combine equation I and II,

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

Conclusion:

Substitute $135.0\text{ MeV}$ for $m_{\pi }{c}^2$ and $938.3\text{ MeV}$ for $m_p{c}^2$ in equation III.

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

Substitute $0.134$ for $\frac{u}{c}$ in above equation to get energy of photon.

    $E_{\gamma }=127\text{ MeV}$

Therefore, the minimum photon energy required is $\text{127 MeV}$.

(b)

To determine

Find the wavelength of the photon at the peak of blackbody spectrum.

Answer to Problem 70CP

The wavelength of the photon at the peak of blackbody spectrum is $1.06\text{ mm}$. 

Explanation of Solution

Write the equation for peak wavelength.

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

Here, ${\lambda }_{\max }$ is the peak wavelength and $T$ is the temperature.

Conclusion:

Substitute $2.73\text{ K}$ for $T$ in equation IV.

    $\begin{array}{c}{\lambda }_{\max }=\frac{2.898\text{ mm}\cdot \text{K}}{\text{2}\text{.73 K}}\\ =1.06\text{ mm}\end{array}$

Therefore, the wavelength of the photon at the peak of blackbody spectrum is $1.06\text{ mm}$.

(c)

To determine

Find the energy of the photon.

Answer to Problem 70CP

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

Explanation of Solution

Write the equation for energy.

    $E_{\gamma }=hf=\frac{hc}{\lambda }$                                                         (V)

Here, $h$ is the Plank’s constant, $f$ is the frequency and $\lambda$ is the wavelength.

Conclusion:

Substitute $1.06\text{ mm}$ for $\lambda$ and $1 240\text{ eV}\cdot {\text{10}}^{-9}\text{ m}$ for $hc$ in equation V.

    $\begin{array}{c}{E}_{\gamma }=\frac{1 240\text{ eV}\cdot {\text{10}}^{-9}\text{ m}}{\text{1}\text{.06}\times {\text{10}}^{\text{-3}}\text{ m}}\\ =1.17\times {10}^{-3}\text{ eV}\end{array}$

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

(d)

To determine

Find the energy of the photon with reference to Earth reference frame.

Answer to Problem 70CP

The energy of the photon with reference to Earth reference frame is $5.81\times {10}^{19}\text{ eV}$. 

Explanation of Solution

In the primed reference frame, the proton is moving to the right at $\frac{u^{\prime }}{c}=0.134$.

The photon moving to the left with $h{f}^{\prime }=1.27\times {10}^8\text{ eV}$.

In the unprimed frame, $hf=1.17\times {10}^{-3}\text{ eV}$.

From the Doppler effect equation the speed of primed frame is,

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

Conclusion:

Then, the speed of proton is,

    $\begin{array}{c}\frac{u}{c}=\frac{u^{\prime }/c+v/c}{1+u^{\prime }v/c^2}\\ =\frac{0.134+1-1.71\times {10}^{-22}}{1+0.134\left(1-1.71\times {10}^{-22}\right)}\\ =1-1.30\times {10}^{-22}\end{array}$

Then the energy of proton is,

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

Therefore, the energy of the photon with reference to Earth reference frame is $5.81\times {10}^{19}\text{ eV}$.