Question

If the de Broglie wavelength of \( \alpha \)-particle and proton are equal, then the momentum \( P \) and kinetic energy \( E \) are related as:

• A. \( P_{\alpha} = P_{p} \)
• B. \( P_{\alpha} > P_{p} \)
• C. \( E_{\alpha} = E_{p} \)
• D. \( E_{\alpha} > E_{p} \)

Hint:

The de Broglie wavelength is inversely proportional to the momentum, and since the mass of an \( \alpha \)-particle is larger than that of a proton, its momentum will be greater for the same wavelength.

Answer 1

The Correct Option is B

The de Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the equation: \[ \lambda = \frac{h}{p} \] where:
- \( h \) is the Planck's constant,
- \( p \) is the momentum of the particle.
For the de Broglie wavelengths of \( \alpha \)-particles and protons to be equal, their momenta must also be equal. However, we know that the mass of an \( \alpha \)-particle is much greater than the mass of a proton. Since the de Broglie wavelength \( \lambda \) is inversely proportional to the momentum \( p \), the \( \alpha \)-particle will have a greater momentum than the proton for the same wavelength.
Step 1: Relationship between momentum and wavelength. 
For the same de Broglie wavelength, the momentum of a particle is inversely proportional to its mass. Since the mass of the \( \alpha \)-particle is greater, its momentum will be higher than that of a proton. 
Step 2: Conclusion. 
Therefore, \( P_{\alpha} > P_{p} \), which corresponds to option (B). 
Final Answer:} (B) \( P_{\alpha}> P_{p} \)