An electron has a speed of 0.783c. Through what potential difference would the electron need to be accelerated from rest in order to reach this speed? The rest mass energy of an electron is 0.511
MeV)
A) 260 kV
B) 310 kV
C) 360 kV
D) 400 kV

Question

Quizplus · Accepted Answer

To find the potential difference required to accelerate an electron to a speed of 0.783c, we use the concept of relativistic kinetic energy. The total energy (E) of an electron moving at relativistic speeds is given by $E = \gamma mc^2$, where $\gamma$ is the Lorentz factor $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$, $m$ is the rest mass of the electron, and $c$ is the speed of light. The rest mass energy ($mc^2$) of an electron is given as 0.511 MeV. The kinetic energy (KE) is the difference between the total energy and the rest mass energy, $KE = E - mc^2$.Given $v = 0.783c$, we calculate $\gamma$ as:$$\gamma = \frac{1}{\sqrt{1 - (0.783)^2}} \approx 1.620$$The total energy is then:$$E = \gamma mc^2 = 1.620 \times 0.511 \, \text{MeV} \approx 0.827 \, \text{MeV}$$The kinetic energy, which is the energy needed to be supplied by the potential difference, is:$$KE = E - mc^2 = 0.827 \, \text{MeV} - 0.511 \, \text{MeV} \approx 0.316 \, \text{MeV}$$To convert this energy into a potential difference, we use the relation $V = \frac{KE}{e}$, where $e$ is the charge of an electron ($1.602 \times 10^{-19} \, \text{C}$). Since $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$, $0.316 \, \text{MeV} = 0.316 \times 10^6 \, \text{eV}$ is equivalent to $0.316 \times 10^6 \, \text{V}$ or $316 \, \text{kV}$.Therefore, the electron needs to be accelerated through a potential difference of approximately 310 kV to reach a speed of 0.783c.

An Electron Has a Speed of 0