Chapter 32, Problem 46GP

To determine

Distance travelled by protons.

Time taken by proton to reach $1\text{ TeV}$ .

Answer to Problem 46GP

Distance, $\mathrm{d}=2.135\times {10}^9\text{ m}$

Time, $\mathrm{t}=7.11\text{ s}$

Explanation of Solution

Given:

Radius, $\mathrm{R}=1\text{ km}$

Final Energy, ${\text{E}}_2=1\text{ TeV}$

Initial energy, ${\mathrm{E}}_1=150\text{ GeV}$

Potential required to accelerate the proton = $2.5\times {10}^6\text{V}$

Energy required to accelerate proton each revolution,

  $\mathrm{E}\text{'}/\mathrm{r}\mathrm{e}\mathrm{v}$ = $\mathrm{V}\mathrm{q}$

  $\mathrm{E}\text{'}/\mathrm{r}\mathrm{e}\mathrm{v}$ = $2.5\times {10}^6\text{eV}$

Formula used:

  $\mathrm{t}=\frac{\mathrm{d}}{\mathrm{c}}$

  $\mathrm{d}=\mathrm{N}\left(2\mathrm{\pi }\mathrm{R}\right)$

  $\mathrm{N}=\frac{\mathrm{\Delta }\mathrm{E}}{\mathrm{E}\text{'}/\mathrm{r}\mathrm{e}\mathrm{v}}$

Where, $\mathrm{c}$ = speed of light

  $\mathrm{d}$ = distance

  $\mathrm{N}$ = number of revolutions

Calculation:

The number of revolutions is calculated as,

  $\begin{array}{l}\mathrm{N}=\frac{\mathrm{\Delta }\mathrm{E}}{\mathrm{E}\text{'}/\mathrm{r}\mathrm{e}\mathrm{v}}\\ \mathrm{N}=\frac{\left(1\times {10}^{12}-150\times {10}^9\right)}{2.5\times {10}^6}\\ \mathrm{N}=340\times {10}^3\text{ rev}\end{array}$

The distance travelled is calculated as,

  $\begin{array}{l}\mathrm{d}=\mathrm{N}\left(2\mathrm{\pi }\mathrm{R}\right)\\ \mathrm{d}=340\times {10}^3\left(2\mathrm{\pi }\times {10}^3\right)\\ \mathrm{d}=2.135\times {10}^9\text{ m}\end{array}$

The time is calculated as,

  $\begin{array}{l}\mathrm{t}=\frac{\mathrm{d}}{\mathrm{c}}\\ \mathrm{t}=\frac{2.135\times {10}^9}{3\times {10}^8}\\ \mathrm{t}=7.11\text{ s}\end{array}$