Q62PE - College Physics (Urone)

Question

(a) What is the effective accelerating potential for electrons at the Stanford Linear Accelerator, if ꝩ= 1.00 ×10⁵  for them? (b) What is their total energy (nearly the same as kinetic in this case) in GeV?

Step-by-Step Solution

Answer

a. The effective acceleration of the electron is obtained as: V = 5.11×10¹⁰V.

b. The total energy is obtained as: E= 51.1 GeV.

Step 1: Given Data

The Stanford Linear Accelerator of electrons is: ꝩ= 1.00 ×10⁵  

Step 2: Define Relativistic Kinetic Energy

In essence, relativistic kinetic energy describes a particle's kinetic energy as the difference between that energy and its rest mass energy. This equation resembles the non-relativistic kinetic energy statement for low velocities.

Step 3: Evaluating the effective acceleration of the electrons

a. With the help of the equation (\[{\rm{28}}{\rm{.52}}\]), the relativistic kinetic energy is obtained by:

\[K{\rm{}}{E_{rel}}{\rm{}}={\rm{}}(\gamma{\rm{}}-{\rm{}}1){\rm{}}m{c^2}\]……………..(I)

We have the value of \[\gamma \] as the relativistic constant.

The value of \[m{c^2}\] is said to be the rest mass energy of the particle.

With the help of the equation (\[{\rm{28}}{\rm{.43}}\]), the relativistic total energy is obtained by:

\[E{\rm{ }}={\rm{ }}\gamma {\rm{ }}m{c^2}\]……………..(II)

We have the value of \[\gamma \] as the relativistic constant.

The value of \[m{c^2}\] is said to be the rest mass energy of the particle.

The rest energy of the electron then will be:

\[m{\rm{ }}{c^2}{\rm{ }} = {\rm{ }}0.511{\rm{ }}MeV\]

The relativistic factor then will be:

\[\gamma {\rm{  }} = {\rm{ }}1.00{\rm{ }} \times {\rm{ }}{10^5}\]

Then, with the help of the first equation, we get:

\begin{aligned}K{E_{rel{\rm{ }}}} = (\gamma  - 1)m{c^2}\\= \left( {1.00 \times {{10}^5} - 1} \right) \times 0.511MeV\\= 51099.5MeV\\= 5.11 \times {10^4}MeV \end{aligned}

…….(II)

The effective potential of the value \[V\] has to accelerate the electron and it is given by:

\begin{aligned}V = \frac{{K{E_{rel}}}}{e}\\= \frac{{5.11 \times {{10}^4}MeV}}{e}\\= 5.11 \times {10^4}MV\\= 5.11 \times {10^{10}}\;V \end{aligned}

Therefore, the effective acceleration of the electron is: \[5.11 \times {10^{10}}\;V\].

Step 4: Evaluating the total energy

b.With the help of the second equation, we obtain:

\begin{aligned}E = \gamma m{c^2}\\= 1.00 \times {10^5} \times 0.511MeV\\= 0.511 \times {10^5}MeV\\= 51.1 \times {10^3}MeV\\= 51.1GeV \end{aligned}

Therefore, the total energy is 51.1GeV.