The surface of a metal is illuminated with the light of 400 nm. The kinetic 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. 3.09 eV
B. 1.42 eV
C. 1.51 eV
D. 1.68 eV

Answer

B

Explanation

`K_(max) = (hc)/(lambda) - W_(0) or W_(0) = (hc)/(lambda) - K_(max)`
`= (1240)/(400) - 1.68 = 3.10 - 1.68 = 1.42` eV