Chapter 27, Problem 36P

To determine

The energy and momentum of each photon.

Answer to Problem 36P

Solution:

The energy of each photon is $0.511\text{MeV}$ and momentum of photon is zero.

Explanation of Solution

Given:

Electron and positron are moving with a speed of: $3.0\times {10}^5\text{m/s}$

Formula used:

$E=\frac{m_0{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}$

Where $m_0$ is the mass of the electron $9.11\times {10}^{-31}\text{kg}$.

c is the speed of the light $3\times {10}^8\text{m/s}$.

v is the velocity of the particles.

Calculation:

As we know that mass of both electron and positron are same that is $9.11\times {10}^{-31}\text{kg}$.

The energy of the produces photon can be determined by using the formula

$E=\frac{m_0{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}$

As $v<<c\text{or}v=0.001c$

The formula of energy become, $E=m_0{c}^2$

$\begin{array}{l}E=m_0{c}^2=9.11\times {10}^{-31}\text{kg}\times (3\times {10}^8\text{m/s})\\ E=0.511\text{MeV}\end{array}$

So, the energy of each photon become $0.511\text{MeV}$

As both the particles have identical mass and moving with same velocity in opposite direction, before the collision the total momentum of the system was zero and after the collision the momentum still remain zero, the two photons must move with equal speed and have same energy.