Question

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

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

Hint:

For particles accelerated through the same potential, \[ \lambda \propto \frac{1}{\sqrt{mq}} \] So heavier particles with larger charge have smaller de Broglie wavelengths.

Answer 1

The Correct Option is C

Concept: The de Broglie wavelength of a particle is given by \[ \lambda=\frac{h}{p} \] When a charged particle is accelerated through a potential \(V\), \[ \lambda=\frac{h}{\sqrt{2mqV}} \] Thus, \[ \lambda \propto \frac{1}{\sqrt{mq}} \] where \(m\) = mass of the particle \(q\) = charge of the particle.

Step 1: Write the relation for the ratio of wavelengths. \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} \]

Step 2: Substitute particle properties. For proton: \[ m_p=m_p,\quad q_p=e \] For alpha particle: \[ m_\alpha=4m_p,\quad q_\alpha=2e \] \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} \]

Step 3: Simplify. \[ = \sqrt{\frac{1}{8}} \] \[ =\frac{1}{2\sqrt{2}} \] Thus approximately, \[ \lambda_\alpha : \lambda_p = 1:2 \]

Step 4: Final ratio. \[ \boxed{1:2} \]