# Number of photoelectrons liberated

1. Jan 17, 2016

### erisedk

1. The problem statement, all variables and given/known data
A beam of light has two wavelengths λ1 (A°, i.e. angstrom) and λ2 with total intensity of I (W/m2) equally distributed amongst the two wavelengths. The beam falls normally on an area A m2 of a clean metallic surface of work function φ (eV). Assume that there is no loss of light by reflection and that each photon has enough energy to eject one electron. Calculate the number of photoelectrons liberated in 2 seconds.

2. Relevant equations
E = nhc/λ

3. The attempt at a solution
Number of electrons liberated due to light of wavelength λ1 :
I/2×A×2 = n1 hc/λ1
n1 = ( IAλ1 )/hc
Similarly, n2 = ( IAλ2 )/hc
And answer will be n1 + n2
Which is indeed correct.
However, I don't understand something.
Shouldn't the number of photoelectrons ejected be independent of the wavelength of incident light (cos here n ∝ λ)? Isn't that what we've always heard, that number of photoelectrons ejected is only dependent on the intensity of incident light, as long as the light has enough energy to supersede the work function?

2. Jan 17, 2016

### PietKuip

Silly problem. First of all because metallic surfaces are quite reflective (up to their plasma frequency).

And of course there is also a wavelength-dependence in the probability of photoemission.

3. Jan 17, 2016

### erisedk

From what I know, the photoelectric current (or the number of photoelectrons emitted) only depends on the intensity of the incident light. Changing the frequency (or wavelength) of the incident light does not change the photoelectric current, as long as the frequency of the incident light is above the threshold frequency.
In this problem, number of photoelectrons emitted is dependent on the wavelength of the incident light. What part of my argument is wrong?

4. Jan 17, 2016

### PietKuip

Of course the photoelectric current must depend on the color of the light. X-rays go straight through.
Here is a plot of the spectral response of some alkali photocathodes (note the logarithmic scale):
http://psec.uchicago.edu/library/photocathodes/zeke_Bialkali.png

The best these optimized materials can do is a quantum efficiency of about 25 % in the blue part of the spectrum.

Last edited: Jan 17, 2016
5. Jan 17, 2016

### erisedk

Thank you! I get it now. I was confusing and overcomplicating some things relating to saturation currents for different frequencies.