The kinetic energy of an electron, α - particle and a proton are...

The formula to calculate the de Broglie wavelength of a particle is given by

$\lambda = \dfrac{h}{\sqrt{2mK}} . . . \left(1\right)$

Now, the masses of the given particles are interrelated by the following relations:

$\begin{cases}{m}_{e} & \approx & \dfrac{{m}_{p}}{1837} \\ {m}_{\alpha} & = & 4{m}_{p}\end{cases}$

The de Broglie wavelength for electron is given by

$\begin{cases}{\lambda}_{e} & = & \dfrac{h}{\sqrt{2{m}_{e}\left(4K\right)}} \\ = & \dfrac{h}{\sqrt{\dfrac{8}{1837}{m}_{p}K}} . . . \left(2\right)\end{cases}$

The de Broglie wavelength for proton is given by

$\begin{cases}{\lambda}_{p} & = & \dfrac{h}{\sqrt{2{m}_{p}\left(2K\right)}} \\ = & \dfrac{h}{\sqrt{4{m}_{p}K}} . . . \left(3\right)\end{cases}$

And, the de Broglie wavelength of alpha particle is given by

$\begin{cases}{\lambda}_{\alpha} & = & \dfrac{h}{\sqrt{2{m}_{\alpha}K}} \\ = & \dfrac{h}{\sqrt{2\left(4{m}_{p}\right)K}} \\ = & \dfrac{h}{\sqrt{8{m}_{p}K}} . . . \left(4\right)\end{cases}$

From equation (2), (3) and (4), it can be concluded that ${\lambda}_{e} > {\lambda}_{p} > {\lambda}_{\alpha}$.