Concept:When a charged particle is accelerated across a voltage gap (potential difference), the electrical work done on the particle is completely converted into its kinetic energy (assuming starting from rest).
Formula:$$ W = \Delta K.E \implies qV = \frac{1}{2}mv^2 $$
Solution:- Substitute the specific charge of an electron: \( q = e \). The electrical energy given to the electron is \( E = eV \).
- Equate this to the classic formula for kinetic energy: \( eV = \frac{1}{2}mv^2 \).
- Multiply both sides by 2 to clear the fraction: \( 2eV = mv^2 \).
- Divide by mass \( m \): \( v^2 = \frac{2eV}{m} \).
- Take the square root to isolate the maximum velocity \( v \): \( v = \sqrt{\frac{2eV}{m}} \).
Why other options are incorrect:Option C forgets the factor of 2 originating from the \( \frac{1}{2} \) in the kinetic energy equation. Options B and D lack the necessary square root required to solve for \( v \) from \( v^2 \).
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