A proton and \( \alpha \)-particle have same kinetic energy. The de-Broglie wavelength ratio \( \displaystyle\frac{\lambda_{p}}{\lambda_\alpha} \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\sqrt{\frac{m_\alpha}{m_p}} \)
The de Broglie wavelength is \( \lambda = h/p \). Writing the momentum through the kinetic energy K, \( p = \sqrt{2mK} \):
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
The two particles have the same K, so \( \lambda \propto 1/\sqrt{m} \).
\( \displaystyle \frac{\lambda_p}{\lambda_\alpha} = \frac{h/\sqrt{2m_pK}}{h/\sqrt{2m_\alpha K}} = \sqrt{\frac{m_\alpha}{m_p}} \)
With \( m_\alpha \approx 4m_p \) this equals 2: the proton has the longer wavelength, as it is the lighter particle.