Chapter 32, Problem 1P

To determine

The total energy of proton having kinetic energy as $\text{5}.\text{633 GeV}$

Answer to Problem 1P

Solution:

The total energy of the proton with $\text{K}.\text{E 4}.\text{65 GeV is 5}.\text{633 GeV}$

Explanation of Solution

Given:

$\begin{array}{l}m=\text{rest mass of proton}=0.\text{983}/{\text{c}}^{\text{2}}\text{GeV},\\ c=\text{speed of light}=\text{3}\times \text{1}{0}^{\text{8}}\text{m}/\text{s}.\end{array}$

$\text{K}.\text{E}=\text{4}.\text{65 GeV}$

Formula used:

The equation for total energy is given by $E=m{c}^2+K.E$ …………………..I

Calculation:

Now substituting the values in equation I

We get

$\begin{array}{l}E=\frac{0.983}{c^2}\times c^2+4.65\\ E=0.983+4.65\\ E=5.633\text{GeV}\end{array}$