if the de broglie wavelength of an electron is 0.3 nanometre, what is its kinetic energy ?
`[h=6.6xx10^(-34) Js, m=9xx10^(-31)kg, 1eV = 1.6xx10^(-19) J]`

Options

A. 1.68 eV
B. 168 eV
C. 16.8 eV
D. 0.168 eV

Answer

C

Explanation

`lamda = (h)/(sqrt(2mE)) therefore lamda^2 = (h^2)/(2mE)`
`therefore E = (h^2)/(2mlamda^2) = ((6.6xx10^(-34))^2)/(2xx9xx10^(-31)xx(0.3xx10^(-9))^2)`
`therefore E = (6.6xx6.6xx10^(-68))/(18xx3xx3xx10^(-51)xx1.6xx10^(-19)) eV`
`E = 16.8 eV`