Question

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV .

What is the maximum kinetic energy of the photoelectrons when light of wavelength 285 nm falls on the same surface?

Answer 1

In the photoelectric effect, the energy of the incoming photon is used partially to extract the photoelectron from the metal (work function) and the rest is converted into kinetic energy of the photoelectrons:

`hf=\phi + K_(max)` (1)
where
h is the Planck constant
f is the photon frequency

`\phi` is the work function

`K_(max)` is the maximum kinetic energy of the photoelectrons

In the first part of the problem, we have light with wavelength
`\lambda=400 nm=400 \cdot 10^(-9) m`, so the photons have frequency

`f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(400 \cdot 10^(-9) m)=7.5 \cdot 10^(14)Hz`
where c is the speed of light; and therefore the energy of the photons is

`E=hf=(6.6 \cdot 10^(-34)Js)(7.5 \cdot 10^(14) Hz)=5.0 \cdot 10^(-19) J`
Converted into electronvolts (
`1 eV= 1.6 \cdot 10^(-19) J`):

`E=5.0 \cdot 10^(-19) J =3.12 eV`

The maximum kinetic energy of the photoelectrons in this case is

`K_(max)=1.10 eV`, so we can find the work function of the metal by re-arranging eq. (1):

`\phi = hf-K_(max) = 3.12 eV-1.10 eV=2.02 eV`


In the second part of the problem, we have light with wavelength
`\lambda=285 nm=285 \cdot 10^(-9)m`, which corresponds to a frequency of

`f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(285 \cdot 10^(-9)m)=1.05 \cdot 10^(15) Hz`
And so the photons have energy

`E=hf=(6.6 \cdot 10^(-34) Js)(1.05 \cdot 10^(15)Hz)=6.93 \cdot 10^(-19) J`
which corresponds to

`E=6.93 \cdot 10^(-19)J = 4.33 eV`

And if we use again eq.(1), we can now find the new maximum kinetic energy of the photoelectrons:

`K_(max)=hf-\phi=4.33 eV-2.02 eV=2.31 eV`