# Photoelectrons and Kmax

1. Feb 6, 2009

### stickplot

1. The problem statement, all variables and given/known data

A metal gives off photoelectrons with maximum kinetic energy Kmax at the rate of N per second when a beam of light of wavelength λ shines on it. How will Kmax and N change if the intensity (brightness) of light is doubled? Assume 100% efficiency.

2. Relevant equations

hc/wavelength=? then. P/?=

h=planks constant
c= speed of light
p=power
3. The attempt at a solution

kinetic energy will remain the same
n will double

Im not sure how to do this problem
if the brightness doubles, will the frequency double? or the wavelength?
and im not sure how that would affect n per second

so what im basically asking is. Is the intensity of light related to the frequency, and if so, in what way?

Last edited: Feb 6, 2009
2. Feb 7, 2009

### Hootenanny

Staff Emeritus
Light intensity is a measure of the [time-averaged] energy flux and has S.I. units of W.m-2. In other words, light intensity is a measure of the average amount of energy passing through the unit surface. In terms of photons, the intensity of light is directly proportional to the number of photons passing through the unit area per unit time.

Last edited: Feb 7, 2009
3. Feb 8, 2009

### stickplot

ok. so does that mean that the photons would double in number since the brightness doubles, and kmax would remain the same.
is this right?

4. Feb 9, 2009

### Hootenanny

Staff Emeritus