91影视

Given data:

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

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,

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

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$\frac{1}{4}m{c}^2$.