Question

The work function of a material is the minimum energy required to remove an electron. Given the work function of gold is 4.90 eV and that of cesium is 1.90 eV. Calculate the maximum wavelength of light for the photoelectric emission of electrons, for gold and cesium.

Answer 1

Answer

given,

work function of gold = 4.90 eV

work function of cesium = 1.90 eV

using energy equation

   $E = \dfrac{hc}{\lambda}$

h is plank's constant

c is the speed of light

   $\lambda = \dfrac{hc}{E}$

for gold

   $\lambda_g = \dfrac{6.624\times 10^{-34}\times 3\times 10^8}{4.90\times 1.6\times 10^{-19}}$

  $\lambda_g = 2.535\times 10^{-7}\ m$

  $\lambda_g = 0.254\ \mu m$

for cesium

   $\lambda_c = \dfrac{6.624\times 10^{-34}\times 3\times 10^8}{1.90\times 1.6\times 10^{-19}}$

  $\lambda_c = 6.537\times 10^{-7}\ m$

  $\lambda_c = 0.654\ \mu m$