Show that the de-Broglie wavelength A of an electron of energy E is given by the relation :
`lambda = h/sqrt(2mE)`.

Explanation

Let E be energy and p momentum of electron,
So `E = 1/2mv^(2) = (m^(2)v^(2))/(2m) =p^(2)/(2m)`
or `p=sqrt(2mE)`……(i)
Since de-Broglie wavelength associated by an electron is given by:
`lambda = h/p`
Using Eq. (i), we get
`lambda = h/sqrt(2mE)`...........(ii)