If the de Broglie wavelengths for a proton and an \( \alpha \)-particle are equal, then the ratio of their velocities respectively will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4 : 1 \)
The de Broglie wavelength of a particle depends on its momentum.
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{m v} \)
Equal wavelengths mean equal momenta, so \( m_p v_p = m_\alpha v_\alpha \). An \( \alpha \)-particle has four nucleons, so \( m_\alpha = 4\,m_p \).
\( \displaystyle \frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} = \frac{4 m_p}{m_p} = \frac{4}{1} \)
So the lighter proton moves four times as fast.