Consider a double-slit interference experiment employing electrons.If the separa

Question

Consider a double-slit interference experiment employing electrons.If the separation between two very narrow slits is 106 nm and the electron energy is 9 eV, what is the angle at which the first brightinterference fringe is found?
1°

Explanation / Answer

the velocity of electron : (1/2)mV^2 = E => V =(2E/m) = 1.779x10^6 m/s => the momentum of electron : Pe = m*V = 1.689x10^-24 kg.m/s => the wavelength of electron beam (as De Broglie): h/ = Pe =>   = h/Pe =4.095x10^-10 m => the angle at which the first bright interference fringe isfound : dsin = => sin = /d =3.863x10^-3 => = 0.221o

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