What should be the velocity of the electron so that its momentum equals of `4000Å` wavelength photon.

Explanation

de-Broglie wavelength of electron
`lambda = (h)/(P) = (h)/(mv)`
`v = (h)/(m lambda)`
`= (6.6 xx 10^(-34))/(9.11 xx 10^(-31)xx 4000 xx 10^(-10)) = (6.6 xx 10^(-34))/(36.44 xx 10^(-38)) = 0.18112 xx 10^(4)`
`v = 1811 ms^(-1)`