Name _________________________________
Block_______________
Remember the speed of light is 3.00 x 108 m/s
and PlankÕs constant is 6.63 x 10-34 J-s. The equations you need are
c = ln
and E = hn You should write the constants and the
equations on a 3x5 inch card that you can use for the unit test. All of these problems can be solved using the
equations above or using dimensional analysis. Use whichever method makes sense
to you!!!
1. A Bunsen burner is lit. A student has sensory preceptors that allows them to feel the photons emitted by the flame with a wavelength of 2.0 x 10-5 m. Find the A) frequency of these photons and B) the portion of the electromagnetic spectrum they are associated with.
METHOD #1 EQUATIONS c = ln and E = hn
A) Wavelength = l = 2.0 x
10-5 m and speed of light = c = 3.00 x 108 m/s
c = ln
3.00 x 108 m/s = 2.0 x 10-5
m (n)
n = 1.5 x 1013 s-1
or 1.5 x 1013 Hz
B) Look at the electromagnetic
spectrum on page 139. This is in the area called infrared which is also known
as heat. So when you stand next to
the flame you feel these photons and assimilate the information in your brain
and call it warmth.
METHOD #2 DIMENSIONAL ANALYSIS
|
A) Wavelength = l
= 2.0 x 10-5 m and speed of light = c = 3.00 x 108 m/s
? 1/s = 1.5 x 1013
s-1 or 1.5 x 1013 Hz |
2. Another photon is emitted by the flame has an energy of 4.23 x 10-19 Joules. Find the A) frequency, B) wavelength in meters and nanometers, and C) the portion of the spectrum for this photon. How might you detect this photon?
METHOD #1
EQUATIONS c = ln and E = hn
A) Energy = E = 4.23 x 10-19J and speed of light = c = 3.00 x
108 m/s and PlankÕs constant = h = 6.63 x 10-34 J-s
E = hn
4.23 x 10-19J = 6.63 x 10-34 J-s (n)
n = 6.38 x 1014 s-1
or 6.38 x 1014 Hz
B) c = ln
3.00 x 108 m/s = (l)6.38 x 1014
s-1
l = 4.70 x 10-7 m
or 470 nm
C)Look at the electromagnetic spectrum on page 139. This is in the area that
is part of the visible spectrum. It
turns out this wavelength and frequency is associated with the color blue so when
you stand next to the flame you see these photons and assimilate the
information in your brain and call it blue.
METHOD #2 DIMENSIONAL ANALYSIS
|
A) Energy = E = 4.23 x 10-19J and speed of
light = c = 3.00 x 108 m/s and PlankÕs constant = h = 6.63 x 10-34
J-s
? 1/s = 6.38 x 1014 s-1 or 6.38 x
1014 Hz |
|||||||
|
B) Energy = E = 4.23 x 10-19J and speed of
light = c = 3.00 x 108 m/s and PlankÕs constant = h = 6.63 x 10-34
J-s
? m = 4.70 x 10-7 m = 470 nm |
3. Ultraviolet light causes a chemical reaction in your skin and darken the pigments in your skin. If the wavelength of these photons is 200. nanometers, find the A) wavelength in meters B) frequency C) energy of these photons.
METHOD #1 EQUATIONS c = ln and E = hn
A) Wavelength = l = 200. nm
so this is just a conversion question. Move the decimal nine places to the left
so 2.00 x 10-7 m.
B) METHOD #1 EQUATIONS c = ln and E = hn
= hc/l
Wavelength = l = 2.00 x
10-7 m and speed of light = c = 3.00 x 108 m/s
c = ln
3.00 x 108 m/s = 2.0 x 10-7
m (n)
n = 1.5 x 1013 1/s or 1.5 x 1015 Hz
C) METHOD #1 EQUATIONS c = ln and E = hn
= hc/l
Wavelength = l = 2.00 x
10-7 m and speed of light = c = 3.00 x 108 m/s and
frequency = n = 1.5 x 1013 1/s
E = hn
E = 6.63 x 10-34 J-s (1.5
x 1013 1/s)
E = 9.94 x 10-21
J
METHOD #2 DIMENSIONAL ANALYSIS
|
A) 200. nm = ? m
? m = 2.00 x 10-7 m |
|||||||
|
B) speed of light = c = 3.00 x
108 m/s and PlankÕs constant = h = 6.63 x 10-34 J-s
? 1/s = 1.5 x 1013 1/s
or 1.5 x 1015 Hz |
|||||||
|
C) speed of light = c = 3.00 x
108 m/s and PlankÕs constant = h = 6.63 x 10-34 J-s
? 1J = 9.94 x 10-21
J |
4. Explain what the following statements mean.
A) The orbits occupied by electrons are quantized.
Quantized means the energy levels or orbitals have specific amounts of energy they are not continuous.
B) When an atom absorbs a photon the atom goes from the ground state to the excited state.
This is an example of energy being
conserved. The photon is kinetic energy in the form of light, it is absorbed by
the electron and the kinetic energy that is lost by the photon equals the gain
in potential energy of the electron as it moves to a higher energy level.
C) How does an atom create light.
When an electron drops from a high to a low energy level or orbital it loses potential energy. This potential energy is converted into kinetic energy in the form of a photon.
D) What is a photon?
A photon is a particle of light. It is a model that is a part of quantum mechanics.
E) Atom emit line spectra.
This combines the ideas of light being emitted by excited atoms and energy levels are quantized. Since the energy levels are quantized, a specific amount of energy is lost when an electron drops from a high to a low energy level. A photon is released with this specific amount of energy which means only a specific frequency of light is created. This specific frequency creates a line with that color of light.
F) Spectra of different elements are like fingerprints.
Since each element has a different number of protons, each element has energy levels with a different amount of potential energy. For example, an electron dropping from the second to the first energy level. H has one proton and He has two protons. That means that when the electron drops it will release more energy in the He atom because of the increased attractive force of the two protons.
G) What is the electromagnetic spectrum.
These are the different categories of waves based on frequency and wavelength.
H) Draw a picture of a wave and describe the terms, wavelength, frequency, speed, crest, and trough.
I) What is the difference between core electrons and valence electrons?
Valence electrons are in the
outermost energy level. All the other electrons are core electrons.,
J) What happens to the potential energy of an electron as it moves from a low to a high energy level or orbital?
The potential energy of the electron
increases as it moves from a low to a high energy level.
K) How is energy conserved when an atom emits light? How does this relate to kinetic energy and potential energy?
An atom emits light when an electron drops from a higher
to a lower energy level or orbital. As the electron loses potential energy, the
atom releases a photon. The greater the drop in potential energy of the
electron, the greater the energy of the photon, the higher the frequency.
5. What is the underlying principle of quantum mechanics. Illustrate your answer using light and electrons as an example of quantum mechanics.
Quantum mechanics expands the view atoms and electrons to
focus on the wave properties of matter.
These waves are related to the energy associated with the matter. For
instance when considering electrons the wave nature of the electrons determines
the amount of energy the electron possesses. The wave nature of electrons
explains why electrons do not crash into the nucleus. When using the quantum
mechanic view of the atom the electron is a wave not a small particle with a
negative charge. It is hard to conceptualize this model because we cannot
experience the wave nature of big objects with our senses. Think back to the lab activity you did
to find the shape of the object under the board. When you cannot directly use
you senses it is hard to create a model of what you are investigating.
Here are some more problems. All of these problems can be
solved using the equations above or using dimensional analysis. Use whichever
method makes sense to you!!!
6. An excited sodium atom emits light with a wavelength of 592 nm. Find the A) wavelength in meters, B) frequency, C) energy of this photon. What part of the electromagnetic spectrum would this photon fall into?
METHOD #1 EQUATIONS c = ln and E = hn
A) Wavelength = l = 592 nm
so this is just a conversion question. Move the decimal nine places to the left
so 5.92 x 10-7 m.
B) METHOD #1 EQUATIONS c = ln and E = hn
= hc/l
Wavelength = l = 5.92 x
10-7 m and speed of light = c = 3.00 x 108 m/s
c = ln
3.00 x 108 m/s = 5.92 x
10-7 m (n)
n = 5.07 x 1014 1/s or 5.07 x 1014 Hz
C) METHOD #1 EQUATIONS c = ln and E = hn
= hc/l
Wavelength = l = 5.07 x
10-7 m and speed of light = c = 3.00 x 108 m/s and
frequency = n = 5.07 x 1014 1/s
E = hn
E = 6.63 x 10-34 J-s (5.07
x 1014 1/s)
E = 3.36 x 10-19 J
METHOD #2 DIMENSIONAL ANALYSIS
|
A) 592. nm = ? m
? m = 5.92 x 10-7 m |
|||||||
|
B) speed of light = c = 3.00 x
108 m/s and PlankÕs constant = h = 6.63 x 10-34 J-s
? 1/s = 5.07 x 1014
1/s or 5.07 x 1014 Hz |
|||||||
|
C) speed of light = c = 3.00 x
108 m/s and PlankÕs constant = h = 6.63 x 10-34 J-s
? J = 3.36 x 10-19
J |
7. An excited H atom emits light with an energy of 1.63 x 10-18 J. Find the A) frequency, B) wavelength of this photon. What part of the electromagnetic spectrum does this photon belong to?
METHOD #1 EQUATIONS c = ln and E = hn
A) Energy = E = 1.63 x 10-18J
and speed of light = c = 3.00 x 108 m/s and PlankÕs constant = h =
6.63 x 10-34 J-s
E = hn
1.63 x 10-18J = 6.63 x 10-34
J-s (n)
n = 2.46 x 1015 s-1 or 2.46 x 1015
Hz
B) c = ln
3.00 x 108 m/s = (l)2.46 x 1015
s-1
l = 1.22 x 10-7 m or 122 nm
C)Look at the electromagnetic
spectrum on page 139. This is in the area that is part of the X-ray spectrum. These photons are so energetic they
travel right through your body, however they are blocked better by the denser
portions of your body like bones.
METHOD #2 DIMENSIONAL ANALYSIS
|
A) Energy = E = 1.63 x 10-18J
and speed of light = c = 3.00 x 108 m/s and PlankÕs constant = h =
6.63 x 10-34 J-s
? 1/s = 2.46 x 1015
s-1 or 2.46 x 1015 Hz |
|||||||
|
B) Energy = E = 1.63 x 10-18J
and speed of light = c = 3.00 x 108 m/s and PlankÕs constant = h =
6.63 x 10-34 J-s
? m = 1.22 x 10-7 m
= 122 nm |
8. Suppose an electron ÒdropsÓ from the fourth to the second energy level of an atom. How would the energy of the photon released compare to the energy of a photon created when the electron dropped from the third to the first energy level? Explain.
Remember the model of electronic
structure in which the distance between the first and second energy level is
greater than the distance from the second to the third, the third to the fourth
is closer still. So an electron
that drops from the fourth to the second energy level loses less potential
energy than the electron that drops from the third to the first. So a higher
energy photon is released when the electron drops from the third to the first
energy level.