The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is `5 eV`. The de Broglie wavelength of the electron is-

Options

A. `(6.6 xx 10^-9)/(sqrt(1456)) m`
B. `(6.6 xx 10^-9)/(sqrt(145.6)) m`
C. `(6.6 xx 10^-9)/(sqrt(1664))`
D. `(6.6 xx 10^-9)/(sqrt(166.4))m`

Answer

B

Explanation

(b) `KE of overline (e)^(-)` ejected = Energy of incident quantum - threshold energy `= 5 eV`
`lamda = (h)/(mV) = (h)/(sqrt(2 m K.E))`
=`(6.6 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 5 xx 1.6 xx 10^-19))`
`m = (6.6)/(sqrt(145.6)) m`.