Question

In an electron gun, the potential difference between the filament and plate is 3000V. What will be the velocity of electron emitting from the gun?

• A. \(3\times10^8\,\text{m/s}\)
• B. \(3.18\times10^7\,\text{m/s}\)
• C. \(3.52\times10^7\,\text{m/s}\)
• D. 3.26×10⁷m/s

Hint:

Electron speed from potential difference uses energy conservation.

Answer 1

The Correct Option is D

Step 1: Energy gained: eV=(1)/(2)mv²
Step 2: v=√((2eV)/(m)) =√frac2(1.6×10⁻19)(3000)9.1×10⁻31 ≈ 3.26×10⁷m/s