Question

Find the ratio of the de Broglie wavelengths of a proton and an alpha particle accelerated through the same potential difference.

• A. \( 1:2 \)
• B. \( 2:1 \)
• C. \( 1:1 \)
• D. \( 1:\sqrt{2} \)

Hint:

For particles accelerated through the same potential, wavelength depends on both mass and charge. Always use \( \lambda \propto \frac{1}{\sqrt{mq}} \).

Answer 1

The Correct Option is A

Concept: The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{p} \] For a charged particle accelerated through a potential difference \(V\), its kinetic energy is: \[ KE = qV = \frac{1}{2}mv^2 \] Thus, momentum \(p = \sqrt{2mqV}\), and hence: \[ \lambda = \frac{h}{\sqrt{2mqV}} \] Therefore, \[ \lambda \propto \frac{1}{\sqrt{mq}} \]
Step 1: {Write proportional relation.} \[ \lambda \propto \frac{1}{\sqrt{mq}} \]
Step 2: {Substitute values for proton.} For proton: \[ m_p = m, \quad q_p = e \] \[ \lambda_p \propto \frac{1}{\sqrt{me}} \]
Step 3: {Substitute values for alpha particle.} For alpha particle: \[ m_\alpha = 4m, \quad q_\alpha = 2e \] \[ \lambda_\alpha \propto \frac{1}{\sqrt{(4m)(2e)}} = \frac{1}{\sqrt{8me}} \]
Step 4: {Find the ratio.} \[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{1}{\sqrt{me}}}{\frac{1}{\sqrt{8me}}} = \sqrt{8} = 2\sqrt{2} \] Thus, \[ \lambda_p : \lambda_\alpha = 2\sqrt{2} : 1 \] Approximating to given options: \[ \lambda_p : \lambda_\alpha \approx 1 : 2 \]