- #1

baseballfan_ny

- 92

- 23

- Homework Statement
- In the photoelectric effect, photoelectrons sometimes leave the surface at es-

sentially the instant that the light is turned on. This is in contrast to the classical picture, in

which it would take a certain amount of time for enough energy to have been accumulated

at the surface of the metal. Suppose that a very weak beam of light of wavelength ##\lambda## and power W were to land on a piece of metal of work function ##\phi##, starting at time t = 0. The first photoelectron is observed at a time t = ##\tau## . How small must ##\tau## be if the measurement is to be incompatible with the classical picture of light?

- Relevant Equations
- ##KE_{max} = E_{incoming} - \phi##

##E_{incoming} = \frac {hc} {\lambda}##

##P = \frac {Work} {t} ##

So I'm kind of confused about how to interpret the question and the idea of there being a small enough ##\tau## for the classical picture to break down.

I started with the max KE eqn: ##KE_{max} = E_{incoming} - \phi##

I suppose ##E_{incoming}## is the power (W) times time and that is also equal to the Planck energy...

## E_{incoming} =W*\tau = \frac {hc} {\lambda}##

And then I just solved for ##\tau##.

##\tau = \frac {hc} {W\lambda}##.

Is this even close? I feel like I'm really missing something here.

I started with the max KE eqn: ##KE_{max} = E_{incoming} - \phi##

I suppose ##E_{incoming}## is the power (W) times time and that is also equal to the Planck energy...

## E_{incoming} =W*\tau = \frac {hc} {\lambda}##

And then I just solved for ##\tau##.

##\tau = \frac {hc} {W\lambda}##.

Is this even close? I feel like I'm really missing something here.