Find the wavelength of an electron having kinetic energy 10 eV. `(h = 6.33 xx 10^(-34) J.s, m_(e) = 9 xx 10^(-31) kg)`

Explanation

Kinetic energy of the electron,
`E = (1)/(mv^(2) = 10eV = 10 xx (1.6 xx 10^(-19))J`
`:. M^(2)v^(2) = 2mE`, or, momentum `p = mv = sqrt(2mE)`
`:.` The de Broglie wavelength of the electron,
`lamda = (h)/(sqrt(2mE)) = (6.63 xx 10^(-34))/([2 xx (9 xx 10^(-31)) xx (10 xx 1.6 xx 10^(-19))]^(1//2))`
`= 3.9 xx 10^(-10) m = 3.9 Å`