Discussion Forum

1)

The surface of a metal is illuminated with a 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 =  1240eV.nm)

Answer:

Option B

Explanation:

Energy of incident light, E = $\frac{12400}{\lambda (Å)}$

E = $\frac{1240}{400}$ = 3.1 eV

E =  W₀ + K_max

 3.1 = W₀+ 1.68 = W₀= 1.42 eV