Question

The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio \( \left( \frac{\lambda_\alpha}{\lambda_p} \right) \) of de Broglie wavelengths associated with them will be:

• A. \( \frac{1}{16} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{2} \)

Hint:

The de Broglie wavelength is inversely proportional to the momentum of the particle. Since the kinetic energy is proportional to the square of the momentum, the de Broglie wavelength will be smaller for particles with higher kinetic energy.

Answer 1

The Correct Option is C

The de Broglie wavelength \( \lambda \) of a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the particle's momentum. Momentum is related to kinetic energy \( K \) and mass \( m \) by \( K = \frac{p^2}{2m} \), which implies \( p = \sqrt{2mK} \). Substituting this into the de Broglie wavelength equation yields \( \lambda = \frac{h}{\sqrt{2mK}} \). Given that the kinetic energy of an alpha particle is four times that of a proton, \( K_\alpha = 4K_p \). The ratio of their de Broglie wavelengths is: \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha K_\alpha}}}{\frac{h}{\sqrt{2m_p K_p}}} = \frac{\sqrt{2m_p K_p}}{\sqrt{2m_\alpha K_\alpha}} = \frac{\sqrt{m_p K_p}}{\sqrt{m_\alpha K_\alpha}} \] Using the relationships \( m_\alpha = 4m_p \) and \( K_\alpha = 4K_p \): \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\sqrt{m_p \cdot K_p}}{\sqrt{4m_p \cdot 4K_p}} = \frac{\sqrt{m_p K_p}}{\sqrt{16 m_p K_p}} = \frac{1}{4} \] Therefore, the ratio of the de Broglie wavelengths is \( \frac{1}{4} \), corresponding to option (C).