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 related to its momentum \( p \) by the equation: \[ \lambda = \frac{h}{p} \] Where:
- \( h \) is Planck's constant,
- \( p \) is the momentum of the particle. Momentum is related to the kinetic energy \( K \) and mass \( m \) by the equation: \[ K = \frac{p^2}{2m} \] Thus, the momentum can be expressed as: \[ p = \sqrt{2mK} \] The de Broglie wavelength then becomes: \[ \lambda = \frac{h}{\sqrt{2mK}} \] Now, we are told that the kinetic energy of the alpha particle is four times that of the proton: \[ K_\alpha = 4K_p \] The ratio of the de Broglie wavelengths is: \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha K_\alpha}}}{\frac{h}{\sqrt{2m_p K_p}}} \] Simplifying this expression: \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\sqrt{2m_p K_p}}{\sqrt{2m_\alpha K_\alpha}} = \frac{\sqrt{m_p K_p}}{\sqrt{m_\alpha K_\alpha}} \] Since the mass of the alpha particle is four times that of the proton \( m_\alpha = 4m_p \), and \( K_\alpha = 4K_p \), we get: \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\sqrt{m_p \cdot K_p}}{\sqrt{4m_p \cdot 4K_p}} = \frac{\sqrt{m_p K_p}}{\sqrt{16 m_p K_p}} = \frac{1}{4} \] Thus, the ratio of the de Broglie wavelengths is \( \frac{1}{4} \). Therefore, the correct answer is option (C).