The surface of a metal is illuminated with the light of 400 nm. The kinectic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm)

Options

A. 1.51 eV
B. 1.68 eV
C. 3.09 eV
D. 1.42eV

Answer

D

Explanation

`(1)/(2)mv_("max")^(2)=(lambda_(c))/(lambda)-w_(0)`
`w_(0)=(lambda_(c))/(lambda)-(1)/(2)mv_("max")^(2)`
`=(1240)/(400)-1.68=1.41eV`