Question

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is:

• A. \( 1 : 2 \)
• B. \( 1 : 4 \)
• C. \( 2\sqrt{2} : 1 \)
• D. \( 1 : 8 \)

Hint:

For particles accelerated through the same potential, their de Broglie wavelength ratio depends on: \[ \lambda \propto \frac{1}{\sqrt{m}} \] where \( m \) is the particle's mass.

Answer 1

The Correct Option is B

Step 1: de Broglie Wavelength Formula The de Broglie wavelength for a particle accelerated through a potential difference \( V \) is: \[ \lambda = \frac{h}{\sqrt{2m eV}} \] where: - \( h \) is Planck’s constant, - \( m \) is the mass of the particle, - \( eV \) is the kinetic energy gained due to acceleration. Step 2: Applying the Formula to Proton and Alpha Particle For a proton: \[ \lambda_p \propto \frac{1}{\sqrt{m_p}} \] For an alpha particle: - Mass \( m_{\alpha} = 4 m_p \), - Charge \( q_{\alpha} = 2e \). Since they are accelerated through the same potential \( V \), the kinetic energy of the alpha particle is: \[ KE_{\alpha} = 2 eV \] \[ \lambda_{\alpha} \propto \frac{1}{\sqrt{4m_p}} \] \[ = \frac{1}{2} \lambda_p \] Thus, the ratio of de Broglie wavelengths: \[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{1}{4} \] Conclusion Thus, the correct answer is: \[ 1 : 4 \]