The de-Broglie's wavelength is given by λ=mvh, where, h is Planck's constant and v is velocity.
Using relation between momentum and kinetic energy, we have λ=2m(KE)h
Here, kinetic energy of electron is stated as KE=qV.
Then, we have λ=2mqVh
Putting the values of m=9.1×10−31kgandq=1.62×10−19C of electron
We get, λ=V1.22nm.
So, x stand for V.