Chemistry Exploration
Name : ______________________________________________
Developing Ideas Read the previous pages/class notes for background information. Results from the
photoelectric effect experiment are illustrated in the diagram below. In this experiment, when light of a given
frequency and intensity shines on a metal surface, electrons may be emitted.
Figure 1: Selected results of the photoelectric effect.
Dim infra-red light shines on
metal. No electrons are emitted.
Dim red light shines on metal. No
electrons are emitted.
Bright infra-red light shines on
metal. No electrons are emitted.
Bright red light shines on metal.
No electrons are emitted.
e-
e-
Dim blue light shines on metal.
Electrons are emitted.
Bright blue light shines on metal.
Electrons are emitted.
Figure 4: Selected results of the photoelectric effect.
1.
According to the wave theory of light, the energy of radiation depends only on the intensity or wave
amplitude (brightness), not the frequency (what type of light, e.g. red light or green light, visible light or
gamma). According to the particle theory of light states the energy of radiation depend only on the
frequency (or type) of light, not on its intensity (brightness). In the particle theory, intensity corresponds to
the number of light particles.
a.
Given a wave model of light, predict which should have more energy, bright red light or dim blue light.
b.
Using the wave model, how should the energy of bright red light and bright blue light compare?
c.
Given a particle model of light, predict which should have more energy, bright red light or dim blue light.
d. Using the particle model, how should the energy of bright red light and bright blue light compare?
What is the Nature of Light?
1
2.
From the data in Figure 1 above, which is more critical in getting electrons to leave the metal, the brightness
(intensity) of light or the frequency of light (its color)? Explain your choice.
3.
Do these results of the photoelectric effect provide support for the wave theory or particle theory of radiation?
Briefly explain your reasoning.
4.
Einstein explained the photoelectric effect using the idea that light consists of particles of light energy called
photons. Each photon carries a specific energy (in joules, J) given by Planck’s relationship:
E = h
If a photon with energy above a certain minimum or threshold value strikes a material, electrons will be
ejected. This minimum energy required to remove an electron is called the work function, . Work function
values for several elements are given in the table below in units of electron volts (eV), where
1.6022
10-19
1 eV =
J. For example, in order to kick an electron out of boron, you must have a minimum energy
of 4.45 eV. If you have 4.44 eV, that is not enough energy to kick an electron out of boron, if you have 4.48 eV –
it is!
Li
2.9
Be
4.98
B
4.45
C
5.0
Na
2.75
Mg
3.66
Al
4.28
Si
4.85
K
2.30
Ca
2.87
Sc
3.5
Ti
4.33
V
4.3
Cr
4.5
Rb
2.16
Sr
2.59
Y
3.1
Zr
4.05
Nb
4.3
Mo
4.6
Cs
2.14
Ba
2.7
La
3.5
Hf
3.9
Ta
4.25
W
4.55
a.
Mn
4.1
Re
5.75
Fe
4.7
Co
5.0
Ni
5.15
Cu
4.6
Zn
4.9
Ga
4.2
Ge
5.0
As
3.75
Se
5.9
Ru
4.71
Rh
4.98
Pd
5.1
Ag
4.6
Cd
4.2
In
4.12
Sn
4.4
Sb
4.6
Te
4.95
Os
4.8
Ir
5.7
Pt
5.65
Au
5.35
Hg
4.49
Tl
3.8
Pb
4.25
Bi
4.2
What general trends do you observe in  as a function of the location of the element in the periodic table?
(For example, look down a column or across a row.)
The general trend down a column is that 
increases
decreases
The general trend across a row (left to right) is that  increases
decreases
b.
If the energy of a blue photon is 4.42  10-19 J, how many electron volts (eV) is that? If the numbers on the
table are representative of the amount of energy that is needed in order to kick out an electron from the
substance, which elements will lose an electron when hit with blue light?
What is the Nature of Light?
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