Answer
$electron \gt neutron \gt alpha~particle$
Work Step by Step
We can use Equation (38-20) to find an expression for the de Broglie wavelength:
$K = \frac{h^2}{2m \lambda^2}$
$\lambda^2 = \frac{h^2}{2m~K}$
$\lambda = \sqrt{\frac{h^2}{2m~K}}$
It is given that the kinetic energy is the same for all three particles.
If a particle has a smaller mass $m$, then it will have a greater de Broglie wavelength.
We can rank the particles in order of their de Broglie wavelengths:
$electron \gt neutron \gt alpha~particle$
