### Select your language

Suggested languages for you:

Americas

Europe

Q17E

Expert-verified
Found in: Page 556

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# The electron mentioned in Section 12.3 for deep inelastic scattering experiments is ${20}{\text{GeV}}$ , and the momentum is given as .${20}{\text{\hspace{0.17em}GeV}}{/}{\text{c}}$ Why so simple a conversion?

The momentum is given as is proved as the conversion is simple.$20\text{\hspace{0.17em}GeV}$

See the step by step solution

## Step 2: Concept of Energy momentum relationship

The energy of the photon in terms of the speed of light ${\text{c}}$ and momentum ${\text{p}}$ is given as,

.${\text{E​\hspace{0.17em}=\hspace{0.17em}pc}}$

The relation between the electron's energy and its momentum is so simple is because the electron is moving so fast that we can consider it is a relativistic particle.

Actually, it is moving so fast that its energy from its rest mass is negligible, and we can use the energy-momentum relationship for massless particles, which is:

${\text{E​\hspace{0.17em}=\hspace{0.17em}pc}}$

.

## Step 3: Explanation of free electron with  20 GeV

Rearrange the above equation for $p$.

$p=\frac{E}{c}$

Thus, an electron with $20\text{\hspace{0.17em}GeV}$ then its momentum is given as, $p=\frac{20\text{\hspace{0.17em}GeV}}{c}$ .

The momentum is given as $P=\frac{20\text{\hspace{0.17em}GeV}}{c}$ is proved as the conversion is simple.

## Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.