Certainly! Let's go through the detailed steps to solve this problem, given the question and result specifications.
### Problem:
An electron starts from rest and falls through a potential rise of 80 V. What is the final speed of the electron?
### Concepts:
1. Electron Charge (q): -1.6 x 10^-19 C
2. Mass of Electron (m): 9.11 x 10^-31 kg
3. Potential Difference (V): 80 V
4. Potential Energy Lost (PE lost): This energy is converted into kinetic energy (KE gained)
### Step-by-Step Solution:
1. Calculate the Potential Energy Lost:
The potential energy lost by the electron when it falls through a potential rise is given by the product of the charge and the potential difference:
[tex]\[
\text{PE}_{\text{lost}} = q \cdot V
\][/tex]
Since the electron is negatively charged, the potential energy lost (in magnitude) is:
[tex]\[
\text{PE}_{\text{lost}} = \left| q \cdot V \right| = \left| (-1.6 \times 10^{-19} \, \mathrm{C}) \times 80 \, \mathrm{V} \right| = 1.28 \times 10^{-17} \, \mathrm{J}
\][/tex]
2. Convert the Lost Potential Energy to Kinetic Energy:
From the conservation of energy principle, the potential energy lost is converted into kinetic energy gained. Thus,
[tex]\[
\text{KE}_{\text{gained}} = \text{PE}_{\text{lost}} = 1.28 \times 10^{-17} \, \mathrm{J}
\][/tex]
3. Express the Kinetic Energy in Terms of Final Speed:
The kinetic energy (KE) can be expressed in terms of the mass (m) and final speed (v) of the electron:
[tex]\[
\text{KE} = \frac{1}{2} m v^2
\][/tex]
4. Solve for Final Speed (v):
Rearrange the kinetic energy formula to solve for the final speed (v):
[tex]\[
\frac{1}{2} m v^2 = \text{KE}
\][/tex]
[tex]\[
v^2 = \frac{2 \cdot \text{KE}}{m}
\][/tex]
[tex]\[
v = \sqrt{\frac{2 \cdot \text{KE}}{m}}
\][/tex]
Plug in the known values:
[tex]\[
v = \sqrt{\frac{2 \cdot 1.28 \times 10^{-17} \, \mathrm{J}}{9.11 \times 10^{-31} \, \mathrm{kg}}}
\][/tex]
Simplify inside the square root:
[tex]\[
v = \sqrt{\frac{2.56 \times 10^{-17} \, \mathrm{J}}{9.11 \times 10^{-31} \, \mathrm{kg}}}
\][/tex]
5. Calculate the Final Speed:
[tex]\[
v \approx \sqrt{2.81 \times 10^{13}} \, \mathrm{m/s}
\][/tex]
[tex]\[
v \approx 5301036.495 \, \mathrm{m/s}
\][/tex]
### Final Answer:
The final speed of the electron after falling through a potential difference of 80 V is approximately [tex]\( 5.3 \times 10^6 \, \text{m/s} \)[/tex].
