Question

A proton and an alpha particle are accelerated through the same potential difference. The ratio of their kinetic energies will be:

• A. \(1:1\)
• B. \(1:2\)
• C. \(2:1\)
• D. \(4:1\)

Hint:

Remember: \[ K=qV \] Greater charge gains greater kinetic energy in same potential difference.

Answer 1

The Correct Option is B

Step 1: Recall kinetic energy gained in electric field. \[ K=qV \] where: \[ q=\text{charge}, \qquad V=\text{potential difference} \]

Step 2: Calculate energies. For proton: \[ K_p=eV \] For alpha particle: \[ K_\alpha=2eV \] Thus: \[ K_p:K_\alpha = 1:2 \] Therefore: \[ \boxed{1:2} \]