Chapter 3, Problem 55P

(a)

To determine

The minimum photon energy required to produce a proton-antiproton pair from collision of photon with a free electron at rest.

Answer to Problem 55P

The minimum required energy for production of electron-positron pair is $2.04\text{MeV}$.

Explanation of Solution

Write the expression for the energy of the electron-positron pair in inertial frame.

  $hf=\sqrt{E_e^2-E_0^2}$        (I)

Here, $h$ is the Planck’s constant, $f$ is the frequency of the photon, $E_e$ is the energy of electron and $E_0$ is the rest energy of the electron.

The rest energy for an electron is $E_0$ and the kinetic energy is $\frac{E_0}{2}$ whereas for the positron, the rest energy and the kinetic energy is $E_0$ and $\frac{E_0}{2}$. Therefore, the total energy possessed by the electron positron pair is $3E_0$.

Write the expression for the initial energy of the electron-positron pair using conservation of energy.

  $hf+E_e=E_t$

Substitute $\sqrt{E_e^2-E_0^2}$ for $hf$ and $3E_0$ for $E_t$ in above expression.

  $\sqrt{E_e^2-E_0^2}=E_e+E_t$        (II)

Here, $E_t$ is the total energy possessed by the pair.

Write the expression for the energy of the electron in the relativistic frame of reference.

  $E_e=\frac{E_0}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$        (III)

Here, $v$ is the speed of the electron and $c$ is the speed of light.

Write the expression for the Doppler’s shift in the energy of the photon in laboratory frame.

  $E_{lab}=hf\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$        (IV)

Here, $E_{lab}$ is the energy in the laboratory frame of reference.

Conclusion:

Substitute $3E_0$ for $E_t$ in equation (II).

  $\begin{array}{c}\sqrt{E_e^2-E_0^2}=E_e+3E_0\\ 6E_e=10E_0\\ E_e=\frac{5}{3}{E}_0\end{array}$

Substitute $\frac{5}{3}{E}_0$ for $E_e$ in equation (III).

  $\begin{array}{c}\frac{5}{3}{E}_0=\frac{E_0}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ \frac{v}{c}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ v=0.8c\end{array}$

Substitute  $\frac{5}{3}{E}_0$ for $E_e$ in equation (I).

  $\begin{array}{c}hf=\sqrt{{\left(\frac{5}{3}{E}_0\right)}^2-E_0^2}\\ =\frac{4}{3}{E}_0\end{array}$

Substitute $\frac{4}{3}{E}_0$ for $hf$, $0.51\text{MeV}$ for $E_0$ and $0.8c$ for $v$ in equation (IV).

  $\begin{array}{c}{E}_{lab}=\left(\frac{4}{3}{E}_0\right)\sqrt{\frac{1+\frac{0.8c}{c}}{1-\frac{0.8c}{c}}}\\ =4\left(0.51\text{MeV}\right)\\ =2.04\text{MeV}\end{array}$

Thus, the minimum required energy for production of electron-positron pair is $2.04\text{MeV}$.

(b)

To determine

The minimum photon energy required to produce a proton-antiproton pair from collision of photon with a free proton at rest.

Answer to Problem 55P

The required minimum energy from a proton at rest is $1.02\text{MeV}$.

Explanation of Solution

Write the expression for the proton’s rest energy.

  $E_p=m{c}^2$        (V)

Here, $E_p$ is the rest energy of the proton and $m$ is the mass of proton.

Write the expression for the energy of the electron-positron pair in inertial frame.

  $hf=\sqrt{E_p^2-{\left(m{c}^2\right)}^2}$

Write the expression for the conservation of the energy for the photon.

  $hf+E_p=2E_0+m{c}^2$        (VI)

Substitute $\sqrt{E_p^2-{\left(m{c}^2\right)}^2}$ for $hf$ in above expression and simplify for $E_p$.

  $\begin{array}{c}\sqrt{E_p^2-{\left(m{c}^2\right)}^2}=2E_0+m{c}^2-E_p\\ E_p=\frac{2E_0^2+{\left(m{c}^2\right)}^2+2E_{0}m{c}^2}{2E_0+m{c}^2}\end{array}$

The value of rest energy of the electron is $0.51\text{eV}$. This amount of energy is very small and it can be neglected in the calculations.

So, the energy of the proton is $E_p=m{c}^2$.

Conclusion:

Substitute $m{c}^2$ for $E_p$ in equation (VI).

  $\begin{array}{c}hf+m{c}^2=2\left(m{c}^2\right)+m{c}^2\\ hf=2E_0\end{array}$

Therefore, the minimum energy required from the rest photon is $2E_0$.

Substitute $0.51\text{MeV}$ for $E_0$ in above expression.

  $\begin{array}{c}E=2\left(0.51\text{MeV}\right)\\ =1.02\text{MeV}\end{array}$

Thus, the required minimum energy from a proton at rest is $1.02\text{MeV}$.