The kinetic energy of an electron, $\alpha$-particle and a proton are given as 4K, 2K and K respectively.  The de-Broglie wavelength associated with electron $\left({\lambda }_e\right),\alpha$-particle $\left({\lambda }_{\alpha }\right)$ and the proton $\left({\lambda }_p\right)$are as follows :

Given that kinetic energy of electron, $\alpha$-particle, proton are respectively 4K, 2K and K.

de-Broglie wavelength of a particle is given by $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$

Where h= Planck’s constant

p = momentum of particle

We know that $\mathrm{p}=\sqrt{2\mathrm{mKE}}$

Where m= mass of particle , KE = kinetic energy of particle.

Let ${\lambda }_e,{\lambda }_{\alpha },{\lambda }_p$ are de-Broglie wavelengths of electron, $\alpha$-particle, proton respectively.

$\therefore {\lambda }_e=\frac{\mathrm{h}}{{\mathrm{p}}_{\mathrm{e}}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{e}}{\mathrm{KE}}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{e}}\left(4\mathrm{K}\right)}}$

${\mathrm{\lambda }}_{\mathrm{\alpha }}=\frac{\mathrm{h}}{{\mathrm{p}}_{\mathrm{\alpha }}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{KE}}_{\mathrm{\alpha }}}}=\frac{\mathrm{h}}{\sqrt{2\left(4{\mathrm{m}}_{\mathrm{p}}\right)2\mathrm{K}}}$

${\mathrm{\lambda }}_{\mathrm{p}}=\frac{\mathrm{h}}{{\mathrm{p}}_{\mathrm{p}}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{p}}{\mathrm{KE}}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2\left({\mathrm{m}}_{\mathrm{p}}\right)\mathrm{K}}}$

$\therefore {\mathrm{\lambda }}_{\mathrm{p}}>{\mathrm{\lambda }}_{\mathrm{\alpha }}$

As, ${\mathrm{m}}_{\mathrm{e}}<<{\mathrm{m}}_{\mathrm{p}};{\mathrm{\lambda }}_{\mathrm{e}}>{\mathrm{\lambda }}_{\mathrm{p}}$

 $\therefore$order of de-Broglie wavelengths is ${\mathrm{\lambda }}_{\mathrm{e}}>{\mathrm{\lambda }}_{\mathrm{p}}>{\mathrm{\lambda }}_{\mathrm{\alpha }}$