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

Calculate the de-Broglie wavelength of an electron that has been accel

1.	Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV.
Answer» Solution :Energy acquired = `1000 eV` `"" = 1000 xx 1.602 xx 10^(-19)` `= 1.602 xx 10^(-16) J` `1/2 mv^2 = 1.602 xx 10^(-16) J` `1/2 xx 9.1 xx 10^(-31) V^2 = 1.602 xx 10^(-16) J` `v^2 = (2 xx 1.602 xx 10^(-16))/(9.1 xx 10^(-31)) = 3.521 xx 10^(14)` `v = 1.876 xx 10^7 m s^(-1)` Now, `lambda = h/(mv)` `= (6.626 xx 10^(-34) kg m^2 s^(-1))/((9.1 xx 10^(-31) kg) xx (1.876 xx 10^(7) ms^(-1)))` `= 3.88 xx 10^(-11) m`