Question

the maximum wavelength of light that can cause photoelectric effect in cesium?

Answer 1

work function for a sample of cesium is 3.43 * 10219 J. (a) What is the minimum frequency of light that will result in
electrons being ejected from this sample by the photoelectric effect? (b) What is the maximum wavelength of light that will
result in electrons being ejected from this sample by the photoelectric effect?
Set Up
Equation 26-1 tells us that the minimum energy
required to eject an electron corresponds to
having Kmax = 0, so the electrons just barely
make it out of the cesium. We’ll use this to
find the threshold frequency f0 that just barely
allows an electron to be ejected. We’ll find the
corresponding wavelength using Equation 22-2.
Maximum kinetic energy of
an electron in the photoelectric
effect:
Kmax = hf 2 0 (26-1)
Propagation speed, frequency, and
wavelength of an electromagnetic
wave:
c = fl (22-2)
Solve
(a) Use Equation 26-1 to calculate the
frequency f that corresponds to Kmax = 0.
From Equation 26-1 with Kmax = 0,
0 = hf 2 0
hf = 0
This says that the photon energy hf is just enough to remove the
most easily dislodged electron from the material (which requires
energy 0), with nothing left over to give the electron any kinetic
energy. So hf = 0 is the minimum photon energy that will eject
an electron, and the frequency f of this photon is the minimum
(threshold) frequency that will do the job:
f0 = f = 0
h
= 3.43 * 10-19 J
6.63 * 10-34 J # s = 5.17 * 1014 s-1 = 5.17 * 1014 Hz
(b) Find the wavelength that corresponds to the
frequency that we calculated in part (a).
We can rewrite Equation 22-2 as
l = c
f
In words, this says that wavelength is inversely proportional to
frequency. So the minimum frequency of light that will eject an
electron corresponds to the maximum wavelength that will eject
an electron:
l max = c
f0
= 3.00 * 108 m>s
5.17 * 1014 Hz
= 5.80 * 1027
m = 580 nm
(Recall that 1 nm = 1029
m