Question

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted electrons is 0.54 eV. What is the maximum kinetic energy if the surface is illuminated using radiation of wavelength 340 nm?

Answer 1

Max kinetic energy for 340 nm wavelength will be
`2.238* 10^(-19)j`

Step-by-step explanation:

In first case wavelength of electromagnetic radiation
`\lambda =480nm=480* 10^(-9)m`

Plank's constant
`h=6.6* 10^(-34)J-s`

Maximum kinetic energy = 0.54 eV

Energy is given by
`E=(hc)/(\lambda )=(6.6* 10^(-34)* 3* 10^8)/(480* 10^(-9))=4.125* 10^(-19)J`

We know that energy is given

`E=K_(MAX)+\Phi`, here
`\Phi` is work function

So
`4.125* 10^(-19)=0.54* 10^(-19)+\Phi`

`\Phi =3.585* 10^(-19)J`

Now wavelength of second radiation = 340 nm

So energy
`E=(hc)/(\lambda )=(6.6* 10^(-34)* 3* 10^8)/(340* 10^(-9))=5.823* 10^(-19)J`

So
`K_(MAX)=5.823* 10^(-19)-3.585* 10^(-19)=2.238* 10^(-19)j`