An electron experiment was performed with a beam of electron accelerated by a potentail difference of `10.0 k eV` .What is the wavelength of the electron beam

Answer

A, B, D

Explanation

`KE = (1)/(2) mv^(2)`
`v = ((2 xx KE)/(m))^((1)/(2))`
`10.0 keV = (10 xx 10^(3) eV)(1.602 xx 10^(-19) J ev^(-1))`
`= 1.602 xx 10^(-15) J = 16.02 xx 10^(-15) kg m^(2)s^(-2)`
`:. V = [(2 xx 1.602 xx 10^(-15) kg m^(2)s^(-2))/(9.1 xx 10^(-31) kg)]^((1)/(2))`
`= (35.17 xx 10^(-14))^((1)/(2)) = 5.93 xx 10^(7) m s^(-1)`
`lambda = (h)/(mv) = (6.63 xx 10^(-34) J s)/((9.1 xx 10^(-31) kg)(5.93 xx10^(7) m s^(-1)))`
`= (1.23 xx 10^(-11) kgm^(2)s^(-1))/(kg m s^(-1))`
`1.23 xx 10^(-11) m = 0.123 Å`