In a pair annihilation experiment, an electron (mass m) with momentum ${\mathbf{p}}_{\mathbf{6}}$hits a positron (same mass, but opposite charge) at rest. They annihilate, producing two photons. (Why couldn’t they produce just one photon?) If one of the photons emerges at $\mathbf{60}\mathbf{°}$to the incident electron direction, what is its energy?

Write the expression for the energy of the electron before the collision.

${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\sqrt{{\mathbf{p}}_{\mathbf{0}}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}^{\mathbf{2}}\mathbf{}{\mathbf{c}}^{\mathbf{4}}}$

Here,${\mathrm{p}}_0$is the initial momentum of the electron, c is the speed of light, and m is the mass of an electron.

Let one photon emerges to the incident electron direction be $\mathrm{\theta }$.

Write the expression for the total energy.

$\mathrm{E}={\mathrm{E}}_1-{\mathrm{E}}_0$ …… (1)

Here,Eis the rest energy of the positron, which is given by,

$\mathrm{E}={\mathrm{mc}}^2$

Substitute ${\mathrm{mc}}^2\mathrm{for}\mathrm{E}\mathrm{and}\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\mathrm{for}{\mathrm{E}}_0$ in equation (1).

$$\begin{gathered} {\mathrm{mc}}^2={\mathrm{E}}_1-\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4} \\ {\mathrm{E}}_1=\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}+{\mathrm{mc}}^2 \end{gathered}$$

Write the expression for the net force after the annihilation (interaction).

${\mathrm{E}}_{\mathrm{t}}={\mathrm{E}}_{\mathrm{A}}+{\mathrm{E}}_{\mathrm{B}}$ …… (2)

Using the law of conservation of energy, the total initial energy of an electron and positron will be equal to the total final energy of the photons

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

Substitute $\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}+{\mathrm{mc}}^2\mathrm{for}{\mathrm{E}}_{\mathrm{i}}\mathrm{and}{\mathrm{E}}_{\mathrm{f}}$in equation (2).

$$\begin{gathered} \sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}+{\mathrm{mc}}^2={\mathrm{E}}_{\mathrm{A}}+{\mathrm{E}}_{\mathrm{B}} \\ {\mathrm{E}}_{\mathrm{B}}=\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}-{\mathrm{mc}}^2-{\mathrm{E}}_{\mathrm{A}} \end{gathered}$$

From the above figure, write the equation for the final momentum of an electron along the horizontal direction.

$$\begin{gathered} {\mathrm{p}}_0=\frac{{\mathrm{E}}_{\mathrm{A}}}{\mathrm{c}}\cos 60°+\frac{{\mathrm{E}}_{\mathrm{B}}}{\mathrm{c}}\mathrm{cos\theta } \\ {\mathrm{p}}_0=\frac{{\mathrm{E}}_{\mathrm{A}}}{\mathrm{c}}\left(\frac{1}{2}\right)+\frac{{\mathrm{E}}_{\mathrm{B}}}{\mathrm{c}}\mathrm{cos\theta } \\ {\mathrm{p}}_0\mathrm{c}=\frac{{\mathrm{E}}_{\mathrm{A}}}{2}+{\mathrm{E}}_{\mathrm{B}}\mathrm{cos\theta } \\ {\mathrm{E}}_{\mathrm{B}}\mathrm{cos\theta }={\mathrm{p}}_0\mathrm{c}-\frac{{\mathrm{E}}_{\mathrm{A}}}{2} \end{gathered}$$ …… (3)

Write the equation for the final momentum of an electron along the vertical direction.

$$\begin{gathered} 0=\frac{{\mathrm{E}}_{\mathrm{A}}}{\mathrm{c}}\sin 60°-\frac{{\mathrm{E}}_{\mathrm{B}}}{\mathrm{c}}\mathrm{sin\theta } \\ 0=\frac{{\mathrm{E}}_{\mathrm{A}}}{\mathrm{c}}\left(\frac{\sqrt{3}}{2}\right)-\frac{{\mathrm{E}}_{\mathrm{B}}}{\mathrm{c}}\mathrm{sin\theta } \\ 0={\mathrm{E}}_{\mathrm{A}}\left(\frac{\sqrt{3}}{2}\right)-{\mathrm{E}}_{\mathrm{B}}\mathrm{sin\theta } \\ {\mathrm{E}}_{\mathrm{B}}\mathrm{sin\theta }=\frac{\sqrt{3}}{2}{\mathrm{E}}_{\mathrm{A}} \end{gathered}$$

Square and add equations (3) and (4).

$$\begin{gathered} {\left({\mathrm{E}}_{\mathrm{B}}\mathrm{sin\theta }\right)}^2+\left({\mathrm{E}}_{\mathrm{B}}\mathrm{cos\theta }\right)={\left(\frac{\sqrt{3}}{2}{\mathrm{E}}_{\mathrm{A}}\right)}^2+{\left({\mathrm{p}}_0\mathrm{c}-\frac{{\mathrm{E}}_{\mathrm{A}}}{2}\right)}^2 \\ {\mathrm{E}}_{\mathrm{B}}^2\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)=\frac{3}{4}{\mathrm{E}}_{\mathrm{A}}^2+\left({\mathrm{p}}_0^2{\mathrm{c}}^2+\frac{{\mathrm{E}}_{\mathrm{A}}^2}{4}-{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}}\right) \\ {\mathrm{E}}_{\mathrm{B}}^2={\mathrm{E}}_{\mathrm{A}}^2+{\mathrm{p}}_0^2{\mathrm{c}}^2-{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}} \end{gathered}$$

Substitute $\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4+{\mathrm{mc}}^2-{\mathrm{E}}_{\mathrm{A}}}\mathrm{for}{\mathrm{E}}_{\mathrm{B}}$in the above expression.

$$\begin{gathered} {\left(\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}+{\mathrm{mc}}^2-{\mathrm{E}}_{\mathrm{A}}\right)}^2={\mathrm{E}}_{\mathrm{A}}^2+{\mathrm{p}}_0^2{\mathrm{c}}^2-{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}} \\ \left[\begin{array}{c}\left({\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4\right)+{\left({\mathrm{mc}}^2-{\mathrm{E}}_{\mathrm{A}}\right)}^2+\\ 2\left(\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\right)\left({\mathrm{mc}}^2-{\mathrm{E}}_{\mathrm{A}}\right)\end{array}\right]={\mathrm{E}}_{\mathrm{A}}^2+{\mathrm{p}}_0^2-{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}} \\ \left[\begin{array}{c}\left({\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4\right)+\left({\mathrm{m}}^2{\mathrm{c}}^4+{\mathrm{E}}_{\mathrm{A}}^2-2{\mathrm{mc}}^2{\mathrm{E}}_{\mathrm{A}}+\right)\\ 2\left(\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\right){\mathrm{mc}}^2-2\left(\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\right){\mathrm{E}}_{\mathrm{A}}\end{array}\right]={\mathrm{E}}_{\mathrm{A}}^2+{\mathrm{p}}_0^2-{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}} \\ -{\mathrm{p}}_0{\mathrm{cE}}_{\mathrm{A}}=\left[\begin{array}{c}2{\mathrm{m}}^2{\mathrm{c}}^4+2{\mathrm{mc}}^2\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}-\\ 2\left(\sqrt{{\mathrm{p}}_0{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\right){\mathrm{E}}_{\mathrm{A}}-2{\mathrm{E}}_{\mathrm{A}}{\mathrm{mc}}^2\end{array}\right] \end{gathered}$$

On further solving,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}}={\mathrm{mc}}^2\left[\frac{{\mathrm{mc}}^2+\sqrt{{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}}{{\mathrm{mc}}^2+\sqrt{{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}-{\displaystyle \frac{1}{2}}{\mathrm{p}}_0\mathrm{c}}\right]\left[\frac{{\mathrm{mc}}^2-\sqrt{{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}-\frac{1}{2}{\mathrm{p}}_0\mathrm{c}}{{\mathrm{mc}}^2-\sqrt{{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}-\frac{1}{2}{\mathrm{p}}_0\mathrm{c}}\right] \\ {\mathrm{E}}_{\mathrm{A}}={\mathrm{mc}}^2\left[\frac{{\mathrm{m}}^2{\mathrm{c}}^4+{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4-\frac{1}{2}{\mathrm{p}}_0{\mathrm{mc}}^3-{\displaystyle \frac{{\mathrm{p}}_0\mathrm{c}}{2}}\sqrt{{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}}{{\mathrm{m}}^2{\mathrm{c}}^4-{\mathrm{p}}_0{\mathrm{mc}}^3+\frac{{\mathrm{p}}_0^2{\mathrm{c}}^2}{2}-{\mathrm{p}}_0^2{\mathrm{c}}^2+{\mathrm{m}}^2{\mathrm{c}}^4}\right] \\ {\mathrm{E}}_{\mathrm{A}}=\frac{{\mathrm{mc}}^2}{2}\left[\frac{\mathrm{mc}+2{\mathrm{p}}_0+\sqrt{{\mathrm{p}}_0^2+{\mathrm{m}}^2{\mathrm{c}}^4}}{\mathrm{mc}+{\displaystyle \frac{3}{4}}{\mathrm{p}}_0}\right] \end{gathered}$$

Therefore, the energy of an electron is${\mathrm{E}}_{\mathrm{A}}=\frac{{\mathrm{mc}}^2}{2}\left[\frac{\mathrm{mc}+2{\mathrm{p}}_0+\sqrt{{\mathrm{p}}_0^2+{\mathrm{m}}^2{\mathrm{c}}^4}}{\mathrm{mc}+{\displaystyle \frac{3}{4}}{\mathrm{p}}_0}\right]$