Answer
$\lambda=\frac{1.226}{\sqrt K}\;nm.eV^{1/2}$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$, ......................$(1)$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle.
Again, using the classical equations for momentum ($p$) and kinetic energy ($K$), the relation between $p$ and kinetic energy $K$ of electron is
$K=\frac{p^2}{2m}$ , where $m$ is the mass of the electron.
Thus, $p=\sqrt {2mK}$ ......................$(2)$
Substituting eq. $2$ in eq. $1$, we get
$\lambda=\frac{h}{\sqrt {2mK}}$ ......................$(3)$
or, $\lambda=\frac{h}{\sqrt {2m}\sqrt K}$
If $K$ is $eV$, then
kinetic energy in electron-volts
$\lambda=\frac{6.63\times 10^{-34}}{\sqrt {2\times 9.1\times 10^{-31}\times\times1.6\times 10^{-19}}\sqrt K}\;m.eV^{1/2}$
or, $\lambda=\frac{1.226\times 10^{-9}}{\sqrt K}\;m.eV^{1/2}$
or, $\lambda=\frac{1.226}{\sqrt K}\;nm.eV^{1/2}$
