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

Short Answer

Expert verified

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

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

Step by step solution

01

Given Data

The Stanford Linear Accelerator of electrons is: 隄= 1.00 脳105

02

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.

03

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:

袄摆碍调袄谤尘调皑皑调贰冲调谤别濒皑皑调袄谤尘调皑皑=调袄谤尘调皑皑(袄驳补尘尘补调袄谤尘调皑皑-调袄谤尘调皑皑1)调袄谤尘调皑皑尘调肠镑2皑袄闭鈥︹赌︹赌︹赌︹赌..(滨)

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:

\[E{\rm{ }}={\rm{ }}\gamma {\rm{ }}m{c^2}\]鈥︹︹︹︹..(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}

鈥︹赌.(滨滨)

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: \[5.11 \times {10^{10}}\;V\].

04

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.

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