Day: December 4, 2021

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202112041209 Homework 1 (Q5)

Suppose that light of intensity 10^{-9}\,\mathrm{W/m^2} normally shines on a metal surface. The metal is made up of a simple cubic lattice of atoms with lattice spacing 0.3\,\mathrm{nm}. Each atom has one free electron. The binding energy at the metal surface is 8\,\mathrm{eV}. Suppose further that the light is uniformly distributed over the surface and all its energy is absorbed by the surface electrons.

(a) If the incident radiation were well described by classical physics, how long would one have to wait after switching on the light source until an electron gains enough energy to be released as a photoelectron?

(b) In reality, how long is this time duration? Explain briefly.

Reading Comprehension.

Highlighting some keyword(s) will help doing the question:

(S1) […] light […] normally shines on a […] surface […] ;

(S2) […] metal is made up of a simple cubic lattice of atoms […] ;

(S3) […] Each atom has one free electron […] .

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2265 Modern Physics Homework 1 Solution.)

(a)

There are eight atoms each lattice and one electron each atom. Hence,

\begin{aligned} t & = \frac{8 \times (1.602\times 10^{-19})}{10^{-9} \times (0.3\times 10^{-9})^2} \\ & = 1.424\times 10^{10}\,\mathrm{s} \\ \end{aligned}

(b)

No electron can be released as a photoelectron, for no photon has energy greater than 8\,\mathrm{eV}. (Why?)