Chemistry Exploration
What is the Nature of Light? – keep for reference
Name : ______________________________________________ turn in pages 4-9 only
Developing Ideas
Read the previous pages/class notes for background information. Results from the photoelectric
effect experiment are illustrated in the diagram on page 6. In this experiment, when light of a
given frequency and intensity shines on a metal surface, electrons may be emitted. The ejected
electrons are measured as current. You may work in small groups/with other students to answer
the following questions.
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) The particle theory of light states the exact opposite! That the
energy of radiation depend only on the frequency of light, not on its intensity (brightness).
Instead, 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. How should the energy of bright red light and bright blue light compare?
According to the wave model, energy depends ONLY on
brightness. Thus bright red light should have more energy than
dim blue light. Since energy depends ONLY on brightness, bright
red light and bright blue light should have the same energy
b. Given a particle model of light, predict which should have more energy, bright red light or
dim blue light. How should the energy of bright red light and bright blue light compare?
According to the particle model of light, the energy depends on the
type of light (e.g. what is its frequency, or what is the wavelength
of the light). According to the particle theory, the higher/larger the
frequency, the higher/larger the energy. Thus, since blue light has
a higher frequency than red light, it should have a higher energy
REGARDLESS if the light is bright or dim – that does NOT matter.
2.
Examine the data in Figure 4 on page 6 and write down at least two observations.
IR light (dim nor bright) cannot eject and electron
Red light (dim nor bright) cannot eject and electron
Blue light: dim OR bright DOES eject an electron
What is the Nature of Light?
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3.
From the data in Figure 4 on page 6, 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.
Since only bright blue light can eject and electron, and since dim
blue light can also eject an electron, it MUST be the frequency and
NOT the brightness that determines if an electron can be ejected.
If it only depended on brightness, all types of bright light should
have been able to eject and electron.
4.
Recall that removing electrons from a metal surface in the photoelectric effect requires light
energy. What does the data in Figure 4 suggest about which is higher energy, red light or
blue light?
BLUE LIGHT!
5.
Do these results of the photoelectric effect provide support for the wave theory or particle
theory of radiation? Briefly explain your reasoning.
This provides support for the particle theory. That the type of light
is more important than the brightness or dimness of the light!
Only blue light –regardless of its intensity (bright or dim!) can eject
an electron
What is the Nature of Light?
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Bright infra-red light shines on
metal. No electrons are emitted.
Dim infra-red light shines on
metal. No electrons are emitted.
Dim 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.
6.
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 = hυ
where h is called Planck’s constant and has the value 6.626 ×10-34 J•s, and υ is frequency in
units of s-1 or Hertz. 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, Φ. Work function values for several elements are given
in the table below in units of electron volts (eV), where 1
eV = 1.6022 ×10-19 J.
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
What is the Nature of Light?
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
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a.
What general trends do you observe in Φ as a function of the location of the element in
the periodic table? (For example, look down a column or across a row.)
Generally speaking, Φ increases as you go from left to right
across the periodic table and it decreases as you go down a
column. This means that it is harder to kick electrons out of the
non-metals and it is easier to kick electrons out of the metals. It
also means that it is harder to kick electrons out of smaller atoms
and easier to kick electrons out of larger atoms (e.g. Li is smaller
than Cs so Li has a larger Φ value) and (Li is larger than C so it
is easier to remove and electron from Li)
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?
1 eV = 1.6022 ×10-19 J.
4.42 x 10-19 J x
1 ev
= 2.75 eV
1.6022 x 10 -19 J
I have 2.75 eV. Any species that has an eV equal to or LESS than
2.75 I can kick the electron out! Na, K, Rb, Sr, Ba, and Cs
AND the following questions: these have to do with lecture – not the information in the handout
above!
1.) A Hydrogen electron makes a transition from n=4 to n=1. Calculate the energy,
frequency, and wavelength of the photon. Do NOT use calculated values to
calculate other numbers unless you absolutely HAVE to! In what region of the
spectrum is this photon?
This energy equation gives you the absolute value of the energy,
regardless if the electron jumped levels or fell from levels.
Therefore, we MUST state, with a sign or words, whether the
energy was absorbed (+E when an electon jumps) or released (-E
when an electron falls) as part of the final answer
⎛ 1
1 ⎞⎟
E = -2.179 x 10-18 J ⎜
−
2
2
⎜n
⎝ H
What is the Nature of Light?
n L ⎟⎠
4
1 ⎞
⎛ 1
− 2⎟
2
1 ⎠
⎝4
E = -2.179 x 10-18 J ⎜
1 ⎞
⎛ 1
− ⎟
16
1 ⎠
⎝
E = -2.179 x 10-18 J ⎜
E = -2.179 x 10-18 J (-0.9375)
E = 2.043 x 10-18 J released or -2.043 x 10-18 J
Now we need to calculate the wavelength, and we SHOULD never use calculated values to
determine other values. If we use the energy to calculate the wavelength, and our energy was
WRONG, our wavelength value would also be incorrect!
1
= 1.0968 x 107 m-1
λ
⎛ 1
1
⎜
−
⎜n 2 n 2
2
⎝ 1
⎞
⎟
⎟
⎠
It really doesn’t matter who is n1 and who is n2 as long as you remember that all wavelengths are
positive numbers – so if you plug in your n values backwards – just take the absolute value of
your answer!!
1
1 ⎞
⎛ 1
= 1.0968 x 107 m-1 ⎜ 2 − 2 ⎟
λ
4 ⎠
⎝1
1
⎛1 1 ⎞
= 1.0968 x 107 m-1 ⎜ −
⎟
λ
⎝ 1 16 ⎠
1
= 1.0968 x 107 m-1 (0.9375)
λ
1
= 1.0283 x 107 m-1
λ
λ=
1
= 9.7248 x 10-8 m (or 97.248 nm)
1.0283 x 10 7 m -1
9.7248 x 10-8 m
Since there is NO other way but to use a calculated value to determine frequency, we MUST use
either wavelength or E. Each should give the same answer. In fact, this is a way to check you
math. If you calculate the frequency based on the wavelength and you get a different number
than the frequency calculated from the energy, somewhere you have made a math mistake! (the
numbers will not be exact due to rounding, but they should be close!!)
What is the Nature of Light?
5
E = hυ
υ = E/h
υ = 2.043 x10-18 J /6.626 x 10-34 Js
υ = 3.083308 x 1015 sec-1
c = λυ
υ = c/λ
υ = 2.9979 x 108m/s /9.7248 x 10-8 m
υ = 3.082736 x 1015 sec-1
Remember a Hertz (Hz is a reciprocal second)
υ from E = 3.083 x 1015 Hz
υ from wavelength = 3.083 x 1015 Hz
Same number!! ☺
2.) Which of the following quantum number sets are allowed? If there is an error
identify AND correct it with a valid quantum number (HELP: the second
number set corresponds to the letter l quantum number.
a.
n= 2,
b. n= 2,
l= 1, ml= 0 allowed
l= 2, ml = 2 not allowed!! largest l value is n-1,
therefore this will work if n > 2 (n = 3, 4, 5, 6,
etc) or if l is smaller and therefore ml must also
be smaller (e.g. l could equal 1 (ml would have to
be -1, 0, or +1) or l could equal 0 and then ml
would have to be zero also!)
c. n= 6,
l= 4,
ml= -5
d. n=6,
l=5,
ml = -4
not allowed!! ml is outside the
range of l value, so ml must be -4, -3, -2, -1, 0, 1,
2, 3, or 4
What is the Nature of Light?
allowed
6
3.) Which of the following represent possible orbitals and which do not – explain
what is wrong with the incorrect symbol
a. 2s:
b. 3p:
possible
possible
not possible!!! There are no 2d orbitals n = 2
l = 2!! Not allowed
c. 2d:
not possible!!! There are no 2f orbitals n = 2
l = 3!! Not allowed
d. 2f:
not possible!!! There are no 1p orbitals n = 1
l = 1!! Not allowed
e. 1p
f.
3d:
g. 8s:
possible
possible someday it will be there!
What is the Nature of Light?
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