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An electron gun has a potential difference of 3000 V between its filament and plate. What speed will an electron acquire as it leaves the gun?
- 3 × 10⁸ m/s
- 3.18 × 10⁷ m/s
- 3.52 × 10⁷ m/s
- 3.26 × 10⁷ m/s
Correct answer: 3.26 × 10⁷ m/s
Solution
v = sqrt(2eV/m) = sqrt(2 * 1.6e-19 * 3000 / 9.11e-31) = 3.25 x 10^7 m/s, which matches 3.26 x 10^7 m/s.
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