# Determining Maximum Wavelength of Incident Radiation

1. Nov 4, 2012

### PHYclass

1. The problem statement, all variables and given/known data
When a certain metallic surface is illuminated with electromagnetic radiation of wavelength λ
max kinetic energy of photoelectrons is 40.5 eV. On same surface infrared radiation of 3λ, the ma kinetic energy is 5% less than that of incident radiation. Determine the maximum wavelength of incident radiation on this surface that will cause photo-emission to occur.

2. Relevant equations

Kmax = hc/λ + ∅

[λ][/c] = hc/∅

3. The attempt at a solution

okay, for wavelength λ I said that

∅= Kmax - hc/λ = 40.5 eV - 1240eV / λ

then for wavelength 3λ

∅= 38.475eV - 1240eV/ 3λ

Now, I want to plug in these values obtained for the work function ∅ into the Max wavelength equation (or cutoff wavelength equation) , but since I have two values what do I do?

Is this the write route

Last edited by a moderator: Nov 5, 2012
2. Nov 5, 2012

### Delphi51

I think "the ma kinetic energy is 5% less than that of incident radiation" means .95*40.5 eV.

If you sub the wavelength and Kmax for each of the two wavelengths into your first formula, you will have two equations with two unknowns, ∅ and λ. If you solve them for the ∅ characteristic of that surface, you will then have an equation relating Kmax to λ with no other unknowns. Use it to find the λ that results in a Kmax of zero. That is the wavelength that just barely causes emission. Any longer wavelength will result in negative Kmax, which is no emission at all.

hc/λ is the energy of the photon. -∅ is the energy required to break an electron out of the metal surface. Kmax is the remaining energy of the electron. That formula is usually written Kmax = hc/λ - W where W is called the Work function of the metal and is the energy required to pop an electron out of the metal.