The equation λ = (1 . 227)/(x)nm can be used to find the de-Broglie...

The de-Broglie's wavelength is given by $\lambda = \dfrac{h}{mv}$, where, $h$ is Planck's constant and $v$ is velocity.

Using relation between momentum and kinetic energy, we have $\lambda = \dfrac{h}{\sqrt{2m\left(KE\right)}}$

Here, kinetic energy of electron is stated as $KE = qV$.

Then, we have $\lambda = \dfrac{h}{\sqrt{2mqV}}$

Putting the values of $m = 9 . 1 \times {10}^{- 31}kgandq = 1 . 62 \times {10}^{- 19}C$ of electron

We get, $\lambda = \dfrac{1 . 22}{\sqrt{V}}nm$.

So, $x$ stand for $\sqrt{V}$.