If numerical value of mass and velocity are equal then De Broglie wave length in terms of K.E is

Given that m = v 

$\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{{\mathrm{m}}^2}---\left(1\right)$

K.E = $\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{m}}^3---\left(2\right)$

multiplying and deviding equation (1) by  "m"

$\mathrm{\lambda }=\frac{\mathrm{hm}}{{\mathrm{m}}^3}---\left(3\right)$

substituting m³ from equation (2) in equation (3) 

$\mathrm{\lambda }=\frac{\mathrm{hm}}{2\mathrm{K}.\mathrm{E}}$