1. Use de Broglie wavelength relation

   The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$ for non-relativistic particles.

2. Given condition

   The proton and the $\alpha$-particle have equal de Broglie wavelengths: $\lambda_p = \lambda_\alpha$

   Therefore, $\frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha}$

   Cancelling $h$, $m_p v_p = m_\alpha v_\alpha$

3. Use mass of the $\alpha$-particle

   An $\alpha$-particle has approximately 4 times the mass of a proton: $m_\alpha = 4m_p$

   Substituting, $m_p v_p = 4m_p v_\alpha$

   Cancelling $m_p$, $v_p = 4v_\alpha$

4. Find the ratio

   Hence, $v_p : v_\alpha = 4 : 1$

5. Option check

   • A: $4:1$ ✅
   • B: $2:1$ ❌
   • C: $1:2$ ❌
   • D: $1:4$ ❌

Therefore, the correct answer is A.