What accelerating potential is needed to produce an electron beam with an effecive wavelength of `0.090 Å`

Answer

`1.86 xx 10^4 V`

Explanation

The kinetic energy of the electron is provided by accelerating potential
`1/2 mv^2 = eV or v = ((2Ve)/(m))^(1/2)`
`:. lamba = h/(mv) = (h)/(m(2Ve//m)^(1//2)) = h/((2 Vem)^(1/2))`
`:. (2 Vem)^(1//2) = h/lambda "or" 2 Vem = (h_2)/(lambda^2)`
or `V = (h^2)/(lambda^2 xx 2 xx e xx m)`
`lambda = 9 pm = 9 xx 10^(-12) m`,
`e = 1.6 xx 10^(-19) C`,
`m = 9.1 xx 10^(-31) kg`
`h = 6.63 xx 10^(-34) Js`
`V = ((6.63 xx 10^(-34)Js)^(2))/((9 xx 10^(-12)m)^(2) xx 2 xx (1.6 xx 10^(-19) C) xx 9.1 xx 10^(-31) kg)`
`= 1.86 xx 10^4 kg m^2 s^(-2)//C = 1.86 xx 10^4 V`.