Question

A neutral pion of (rest) mass mand (relativistic) momentum $\mathit{p}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\mathit{m}\mathit{c}$decays into two photons. One of the photons is emitted in the same direction as the original pion, and the other in the opposite direction. Find the (relativistic) energy of each photon.

Short answer

The energy of photon A is$m{c}^2$, and the energy of photon B is$\frac{1}{4}m{c}^2$.

Step-by-step solution

Given information:

Given data:

The momentum is$p=\frac{3}{4}mc$.

Determine the total energy of neutral pion:

Write the expression for the total energy of pion.

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

Here, p is the momentum, m is the mass, and cis the speed of light.

Substitute $p=\frac{3}{4}mc$in the above expression.

$$\begin{gathered} E=\sqrt{{\left(\frac{3}{4}mc\right)}^2{c}^{2+}{m}^2{c}^4} \\ E=\sqrt{\frac{9}{16}{m}^2{c}^{4+}{m}^2{c}^4} \\ E=\sqrt{\frac{9m^2{c}^{4+}{m}^2{c}^4}{16}} \end{gathered}$$

On further solving,

$$\begin{gathered} E=\sqrt{\frac{25}{16}m{c}^2} \\ E=\frac{5}{4}m{c}^2 \end{gathered}$$

Determine the energy of each photon:

Let the energy of one photon be $E_A$and the energy of another photon be $E_8$.

As the energy is conserved, write the energy equation.

$E=E_A+E_{B......\left(1\right)}$

Write the equation for conservation of momentum.

$P=P_A+P_B...\left(2\right)$

Here, $p_A$is the momentum of one photon and $p_B$is the momentum of another photon.

Write the equation for momentum of one photon.

$p_A=\frac{E_A}{c}$

Write the equation for momentum of another photon.

$p_B=\frac{E_B}{c}$

Here, a negative sign indicates the opposite direction.

Substitute $\frac{5}{4}m{c}^2$for Ein equation (1).

$\frac{5}{4}m{c}^2=E_A+E_B.......\left(3\right)$

Substitute $\frac{3}{4}mc$for $p,\frac{E_A}{c}$for $p_A$and $\frac{E_B}{c}$for $p_B$in equation (2).

$$\begin{gathered} \frac{3}{4}mc=\left(\frac{E_A}{c}\right)+\left(\begin{array}{c}-\frac{E_B}{c}\end{array}\right) \\ {E}_A-E_B=\frac{3}{4}m{c}^2...\left(4\right) \end{gathered}$$

Subtract equation (4) from equation (3).

$$\begin{gathered} E_A+E_B-E_A+E_B=\frac{5}{4}m{c}^2-\frac{3}{4}m{c}^2 \\ 2E_B=\frac{2}{4}m{c}^2 \\ {E}_B=\frac{1}{4}m{c}^2 \end{gathered}$$

Substitute the value of $E_B$in equation (4).

$$\begin{gathered} E_A-\frac{1}{4}m{c}^2=\frac{3}{4}m{c}^2 \\ {E}_A=\frac{3}{4}m{c}^2+\frac{1}{4}m{c}^2 \\ {E}_A=m{c}^2 \end{gathered}$$

Therefore, the energy of photon A is$m{c}^2$and the energy of photon B is$\frac{1}{4}m{c}^2$.