The Question is from neet-physics electrostatic-charges-and-fields , and it was created by the author NAGA SAI

Question

A proton and an alpha particle, both initially at rest, are accelerated through the same potential difference $V$.  If the proton acquires a velocity $v_p$, what is the velocity of the alpha particle?

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: $v_p / \sqrt{2}$ (Option C)
## CONCEPT FRAMEWORK
 **Primary Concept**: Conservation of energy and relationship between kinetic energy and potential difference.
 **Related Topics**: Kinetic energy, potential energy, charge, mass.
 **Prerequisites**: Understanding of potential difference, kinetic energy formula, and charge of particles.
## SOLUTION PATHWAY
**Given Information**:
  * Proton velocity = $v_p$
  * Potential difference for both particles = $V$
  * Initial velocity of both particles = 0
  * Charge of proton ($q_p$) = $+e$
  * Mass of proton ($m_p$) = $m_p$
  * Charge of alpha particle ($q_\alpha$) = $+2e$
  * Mass of alpha particle ($m_\alpha$) = $4m_p$
**Method & Steps**:
 1. Apply conservation of energy.  The potential energy lost is equal to the kinetic energy gained.
  Reasoning:  The potential energy is converted into kinetic energy.
  Formula/Rule: $qV = \frac{1}{2}mv^2$
 2. Apply the formula to both the proton and the alpha particle:
  For proton: $eV = \frac{1}{2}m_pv_p^2$   (1)
  For alpha particle: $2eV = \frac{1}{2}(4m_p)v_\alpha^2$  (2)
 3. Solve for $v_\alpha$ by dividing equation (2) by equation (1):
  $\frac{2eV}{eV} = \frac{\frac{1}{2}(4m_p)v_\alpha^2}{\frac{1}{2}m_pv_p^2}$
  $2 = \frac{4v_\alpha^2}{v_p^2}$
  $v_\alpha^2 = \frac{v_p^2}{2}$
  $v_\alpha = \frac{v_p}{\sqrt{2}}$
  Result: The velocity of the alpha particle is $\frac{v_p}{\sqrt{2}}$
  Verification: The result shows that the alpha particle has a lower velocity than the proton, which is consistent with its higher mass and the same potential difference.
 ## OPTION ANALYSIS
[A] $v_p$: :x: Incorrect. This assumes both particles have the same mass.
[B] $v_p / 2$: :x: Incorrect. This doesn't account for the difference in mass and charge.
[C] $v_p / \sqrt{2}$: ✓: Correct. This accounts for the differences in mass and charge.
[D] $2v_p$: :x: Incorrect. This would imply the alpha particle has a higher velocity, which is not possible given its higher mass.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the difference in mass and charge of the particles: Remember to use the correct mass and charge for each particle in the energy conservation equation.
  2. Incorrectly applying the kinetic energy formula: Ensure the formula is used correctly and consistently for both particles.
 **Quick Tips**:
  1.Use the ratio of the equations to eliminate common terms and simplify the calculation.
  2.Remember that the kinetic energy is proportional to $v^2$ and inversely proportional to mass.

NAGA SAI · Answer

$v_p$

NAGA SAI · Answer

$v_p / 2$

NAGA SAI · Answer

The correct Option is "$v_p / \sqrt{2}$"

NAGA SAI · Answer

$2v_p$

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