An α particle, a proton and an electron have the same kinetic...

As we know, de-Broglie wavelength is given by,

$\lambda = \dfrac{h}{p}$

and $p = \sqrt{2mK}$

Therefore,

$\lambda = \dfrac{h}{\sqrt{2mK}}$

As kinetic energy is the same for all particles,

$∴ \lambda ∝ \dfrac{1}{\sqrt{m}}$

$∵ {m}_{\alpha} > {m}_{p} > {m}_{e}$

${\lambda}_{e} > {\lambda}_{p} > {\lambda}_{\alpha}$