(a) What is the effective accelerating potential for electrons at the Stanford Linear Accelerator, if \(\gamma=1.00 \times 10^{5}\) for them? (b) What is their total energy (nearly the same as kinetic in this case) in GeV?

In order to find the effective accelerating potential, we'll be using the equation that relates energy and potential: \(E = eV\) where \(E\) is the electron energy, \(e\) is the electron charge and \(V\) is the accelerating potential. First, we need to determine the electron energy using its Lorentz factor: \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) Rearrange the formula to solve for energy: \(E = (\gamma - 1)m_ec^2\) where \(\gamma\) is the Lorentz factor (given), \(m_e\) is the electron mass and \(c\) is the speed of light. Now, insert the known values, and solve the equation for the electron energy \(E\): \(E = [1.00 \times 10^{5} - 1] \cdot 9.11 \times 10^{-31}\text{ kg} \cdot (3 \times 10^8\text{ m/s})^2\) Calculate \(E\). Next, use the energy to find the accelerating potential: \(V = \frac{E}{e}\) where \(e\) is the electron charge (\(1.6 \times 10^{-19}\text{ C}\())). Calculate the effective accelerating potential, \(V\).

We already found the total energy \(E\) in part (a). To convert it to GeV, we simply divide by the appropriate conversion factor: \(E_\text{GeV} = \frac{E}{1.6 \times 10^{-10}\text{ Joules/eV}}\) Calculate the total energy of the electrons in GeV.