The kinetic energy of an electron with de - Broglie wavelength of `0.3 nanometre` is

Options

A. `0.168 eV`
B. `16.8 eV`
C. `1.68 eV`
D. `2.5 eV`

Answer

A

Explanation

`lambda = (h)/(sqrt(2 mE)) rArr E = (h^(2))/( 2 m lambda^(2))`
`= ((6.6 xx 10^(-34))^(2))/(2 xx 9.1 xx 10^(-31) xx (0.3 xx 10^(-9))^(2))`
`= 2.65 xx 10^(-18) J = 16.8 eV`