Questions: If an electron is accelerated from rest through a potential difference of 1,500 V, what speed does it reach? 2.3 × 10^7 m / s 1.9 × 10^7 m / s 1.5 × 10^7 m / s 1.1 × 10^7 m / s

If an electron is accelerated from rest through a potential difference of 1,500 V, what speed does it reach?
2.3 × 10^7 m / s
1.9 × 10^7 m / s
1.5 × 10^7 m / s
1.1 × 10^7 m / s

Transcript text: If an electron is accelerated from rest through a potential difference of $1,500 \mathrm{~V}$, what speed does it reach? $2.3 \times 10^{7} \mathrm{~m} / \mathrm{s}$ $1.9 \times 10^{7} \mathrm{~m} / \mathrm{s}$ $1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ $1.1 \times 10^{7} \mathrm{~m} / \mathrm{s}$

Solution

Solution Steps

Step 1: Determine the Energy Gained by the Electron

The energy gained by an electron when it is accelerated through a potential difference $ V $ is given by the equation $ E = eV $, where $ e $ is the charge of the electron ($1.6 \times 10^{-19} \, \text{C}$) and $ V $ is the potential difference.
For a potential difference of $ 1,500 \, \text{V} $, the energy gained is $ E = 1.6 \times 10^{-19} \, \text{C} \times 1,500 \, \text{V} $.

Step 2: Relate Energy to Kinetic Energy

The energy gained by the electron is converted into kinetic energy, which is given by the equation $ \frac{1}{2}mv^2 $, where $ m $ is the mass of the electron ($9.11 \times 10^{-31} \, \text{kg}$) and $ v $ is the speed of the electron.
Set the kinetic energy equal to the energy gained: $ \frac{1}{2}mv^2 = eV $.

Step 3: Solve for the Speed of the Electron

Rearrange the equation to solve for $ v $: $ v = \sqrt{\frac{2eV}{m}} $.
Substitute the known values: $ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \, \text{C} \times 1,500 \, \text{V}}{9.11 \times 10^{-31} \, \text{kg}}} $.
Calculate $ v $ to find the speed of the electron.

Final Answer

$\boxed{2.3 \times 10^{7} \, \text{m/s}}$

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