# An electron with an initial speed of 660,000 m/s is brought to rest by an electric field. A)...

## Question:

An electron with an initial speed of 660,000 m/s is brought to rest by an electric field.

A) What was the potential difference that stopped the electron?

B) What was the initial kinetic energy of the electron, in electron volts?

## Conservation of Energy

When only conservative forces act on a system, the total energy of the system remains conserved, but it can get transformed from one form to another.

Data Given

• The speed of the electron {eq}v= 6.60 \times 10^5 \ \rm m/s {/eq}
• Charge on an electron {eq}q = 1.6 \times 10^{-19} \ \rm C {/eq}
• Mass of an electron {eq}m = 9.11 \times 10^{-31} \ \rm kg {/eq}

Part A) As the electron enters in a region of an electric field, it's kinetic energy starts converting into electrical potential energy. At the complete stop whole of the kinetic energy gets converted into electrical potential energy.

{eq}\begin{align} \frac{1}{2} mv^2 = q \Delta V \end{align} {/eq}

{eq}\begin{align} \Delta V = \frac{mv^2}{2q} \end{align} {/eq}

{eq}\begin{align} \Delta V = \frac{9.11 \times 10^{-31} \ \rm kg \times (6.60 \times 10^{5} \ \rm m/s)^2}{2 \times 1.6 \times 10^{-19} \ \rm C} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{\Delta V = 1.24 \ \rm V }} \end{align} {/eq}

Part B) The initial kinetic energy of the electron is given by

{eq}\begin{align} KE_i = \frac{1}{2} mv^2 \end{align} {/eq}

{eq}\begin{align} KE_i = \frac{1}{2} \times 9.11 \times 10^{-31} \ \rm kg \times (6.60 \times 10^{5} \ \rm m/s)^2 \end{align} {/eq}

{eq}\begin{align} KE_i = 1.984 \times 10^{-19} \ \rm J \times \frac{1}{1.6 \times 10^{-19} \ \rm J/eV} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{KE_i = 1.24 \ \rm eV}} \end{align} {/eq} 