 #1
baseballfan_ny
 91
 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.