Question

The surface of a metal is illuminted with the light of `400 nm` The kinetic energy of the ejection photoelectron was found to be `1.68 eV` The work function of the metal is :
A. `1.41 eV`
B. `1.51 eV`
C. `1.68 eV`
D. `3.09 eV`

Answer 1

Correct Answer - A
`lambda = 400 nm , h c = 1240 eV nm K.E = 1.68 eV`
we know that `(hc)/(lambda) - W = K.E rArr W = (hc)/(lambda) K.E `
`rArr W = (1240)/(400) - 1.68 = 3.1 - 1.68 = 1.42 eV`