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An electron starts from rest and falls through a potential rise of 80 V. What is its final speed?

Positive charges tend to fall through potential drops, while negative charges such as electrons tend to fall through potential rises. The lost potential energy (PE) appears as the kinetic energy (KE) of the electron.

[tex]\[
\text{PE lost} = \text{KE gained}
\][/tex]

Given:
[tex]\[
\text{PE lost} = V \cdot q = (80 \, \text{V}) \left(-1.6 \times 10^{-19} \, \text{C}\right) = -1.28 \times 10^{-17} \, \text{J}
\][/tex]

Since the lost PE equals the gained KE:
[tex]\[
-1.28 \times 10^{-17} \, \text{J} = \frac{1}{2} m v^2
\][/tex]

Solving for [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{\frac{2 \left(1.28 \times 10^{-17} \, \text{J}\right)}{9.1 \times 10^{-31} \, \text{kg}}} = 5.3 \times 10^6 \, \text{m/s}
\][/tex]

The final speed of the electron is [tex]\( 5.3 \times 10^6 \, \text{m/s} \)[/tex].

Sagot :

GinnyAnsweridn
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].
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