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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.

Answer and Explanation:

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}


Learn more about this topic:

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Energy Conservation and Energy Efficiency: Examples and Differences

from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9
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