Question

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?

Short answer

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

Step-by-step solution

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:

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

.

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.