1. For a particle, the de-Broglie wavelength is $\lambda = \frac{h}{p}$ where $h$ is Planck's constant and $p$ is momentum.

2. Since all three particles have the same kinetic energy $K$, use the non-relativistic relation: $K = \frac{p^2}{2m}$ So, $p = \sqrt{2mK}$

3. Substitute into de-Broglie relation: $\lambda = \frac{h}{\sqrt{2mK}}$ Since $h$ and $K$ are same for all three particles, $\lambda \propto \frac{1}{\sqrt{m}}$ Thus, smaller mass means larger wavelength.

4. Compare masses:

• Electron has the smallest mass.
• Proton is much heavier than electron.
• $\alpha$-particle has mass about $4$ times the proton mass, so it is the heaviest.

Therefore, $m_\alpha > m_p > m_e$ Hence, $\lambda_\alpha < \lambda_p < \lambda_e$

5. Check options:

• A: $\lambda_\alpha > \lambda_p < \lambda_e$ ❌
• B: $\lambda_\alpha > \lambda_p > \lambda_e$ ❌
• C: $\lambda_\alpha = \lambda_p = \lambda_e$ ❌
• D: $\lambda_\alpha < \lambda_p < \lambda_e$ ✅

So the correct option is D.