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

Answer

`3.88 xx 10^(-11) m`

Explanation

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`