Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential differecne of 1kV.

Explanation

Energy acquired by the electron (as kinetic energy) after being aacelerated by a potential difference of 1 kV. (i.e. 1000 volts) `= 1000 eV = 1000 xx 1.602 xx 10^(-19) J = 1.602 xx 10^(-16) J (1 eV = 1.602 xx 10^(-19)J)`
i.e., Kinetic energy, `(1)/(2) mv^(2) = 1.602 xx 10^(-16) J or (1)/(2) xx 9.1 xx 10^(-31) v^(2)`
or `v^(2) = 3.521 xx 10^(14) or v = 1.88 xx 10^(7) ms^(-1)`
`:. lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx (1.88 xx 10^(7) ms^(-1))) = 3.87 xx 10^(-11) m`