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?

Short 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-by-step solution

Given Data

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

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.

Evaluating the effective acceleration of the electrons

a. With the help of the equation ($$28.52$$), the relativistic kinetic energy is obtained by:

$$K{E}_{rel}=(\gamma -1)m{c}^2$$……………..(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 ($$28.43$$), the relativistic total energy is obtained by:

$$E=\gamma m{c}^2$$……………..(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:

$$m{c}^2=0.511MeV$$

The relativistic factor then will be:

$$\gamma =1.00\times {10}^5$$

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

$$\begin{array}{r}K{E}_{rel}=(\gamma -1)m{c}^2\\ =\left(1.00\times {10}^5-1\right)\times 0.511MeV\\ =51099.5MeV\\ =5.11\times {10}^{4}MeV\end{array}$$

…….(II)

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

$$\begin{array}{r}V=\frac{K{E}_{rel}}{e}\\ =\frac{5.11\times {10}^{4}MeV}{e}\\ =5.11\times {10}^{4}MV\\ =5.11\times {10}^{10}V\end{array}$$

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

Evaluating the total energy

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

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

Therefore, the total energy is 51.1GeV.