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Chapter 36: Q. 72 (page 1062)

Consider the inelastic collision $e^{-} + e^{-} 鈫� e^{-} + e^{-} + e^{-} + e^{+}$ in which an electron-positron pair is produced in a head-on collision between two electrons moving in opposite directions at the same speed. This is similar to Figure 36.39, but both of the initial electrons are moving.
a. What is the threshold kinetic energy? That is, what minimum kinetic energy must each electron have to allow this process to occur?
b. What is the speed of an electron with this kinetic energy?

Short Answer

Expert verified

a. The minimum energy is $2 m_{e} c^{2} .$

b. the speed of the electron is $\frac{\sqrt{3}}{2} c .$

Step by step solution

01

Part (a) Step 1: Given information

We have given,

$e^{-} + e^{-} 鈫� e^{-} + e^{-} + e^{-} + e^{+}$

We have to find the minimum kinetic energy of the electron.

02

Simplify

This reaction will happens when two electron is moving with some energy and there is happens collision between them and it will produces the electron and positron pair and becomes itself in rest.

Then the energy in the system initially will be,

$E_{i} = 2 m c^{2} + 2 K . E .$

and after the collision will happens the energy will be,

$E_{f} = 4 {mc}^{2}$

From the conservation of energy we can write,

$E_{i} = E_{f} 2 {mc}^{2} + 2 K . E . = 4 {mc}^{2} K . E . = {mc}^{2}$

03

Part (b) Step 1: Given information

We have given,

$e^{-} + e^{-} 鈫� e^{-} + e^{-} + e^{-} + e^{+}$

We have to find the speed of an electron .

04

Simplify

Since,

$K . E . = \frac{{mc}^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} - {mc}^{2} {mc}^{2} = \frac{{mc}^{2}}{^{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} - {mc}^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{1}{2} v = \frac{\sqrt{3}}{2} c$

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