# The Photelectric Effect

1. Feb 20, 2006

### Calmeir

Alright, I am having some trouble with the following problem, because my solution I arrive at seems to be way too large. The problem is:

What is the maximum photelectron speed if potassium is illuminated with light of 250 nm?

From that I have

λ = 250 nm
f = c/λ = 3x10^8 / 250x10^-9 = 1.2x10^15
Eo of potassium = 2.3 eV

From there I tried getting the velocity by obtaining the kinetic energy.

Eelec = hf = (4.14x10^-15)(1.2x10^15) = 4.968
Kmax = Eelec - Eo = 4.968 - 2.3 = 2.668

V = √2K/m = √2(2.668)/(9.11x10^-31) = 2.42x10^15 m/s

So my calculated answer is 2.42x10^15 m/s which is much faster than the speed of light, so I know it must be wrong. I can't seem to figure out another way to arrive at an answer though. Any help is much appreciated.

2. Feb 20, 2006

### inha

Your units don't match. You either need to convert the electrons energy to joules or it's mass to eV/c^2 (that's about 511keV/c^2).
Doing a bit of dimensional analysis is usually useful if you get bizarre results from a calculation.

3. Feb 20, 2006

### Calmeir

Ok, so taking the K of 2.668 eV and converting to joules I got 4.3x10^-19, then plugging those results into the my final equation I ended up with 9.7x10^5 m/s which sounds much more reasonable, but I just want to double check that I did that portion correctly. Was that the result you ended up with?