An electron, an $\alpha$ particle, a proton and a deutron have the same kinetic energy. Which of these particles has the shortest De Broglie wavelength.

$\alpha$ Particle

The correct answer is Option (3) → $\alpha$ Particle

Using De-Broglie,

$λ=\frac{h}{P}=\frac{h}{\sqrt{2mk}}$

$⇒λ∝\frac{1}{\sqrt{m}}$

and, the $\alpha$-Particle has greatest mass ⇒ Shortest wavelength.