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

given,

work function of gold = 4.90 eV

work function of cesium = 1.90 eV

using energy equation

`E = (hc)/(\lambda)`

h is plank's constant

c is the speed of light

`\lambda = (hc)/(E)`

for gold

`\lambda_g = (6.624* 10^(-34)* 3* 10^8)/(4.90* 1.6* 10^(-19))`

`\lambda_g = 2.535* 10^(-7)\ m`

`\lambda_g = 0.254\ \mu m`

for cesium

`\lambda_c = (6.624* 10^(-34)* 3* 10^8)/(1.90* 1.6* 10^(-19))`

`\lambda_c = 6.537* 10^(-7)\ m`

`\lambda_c = 0.654\ \mu m`