When the electron is accelerate through a potential difference of V voilits ,it acquired a kinetic energy by `E = qV` where q is the charge on the electron .Also if m is its mass and v Is its velocity ,then
`E = (1)/(2) mv^(2)`
rArr v = sqrt((2E)/(m))`
And the Broglie wavelength
`lambda = (h)/(mv) = (h)/(sqrt(2Em))`