Chem 31 SI 11/01/09
Chapter 7 Worksheet
Chapter 7 Worksheet
1. A) Draw a basic sine wave
B) Superimpose a second wave with the same wavelength but stronger amplitude
C) Superimpose a third wave with the same amplitude as the first but with a shorter wavelength
A) Blue = Original Wave
B) Pink = Same Wavelength, Stronger Amplitude
C) Green = Same Amplitude as Original, Shorter Wavelength
2. The size of the amplitude corresponds to:
A. The number of wave cycles per second
B. The intensity of the electromagnetic radiation (brightness)
C. The amount of energy contained in an electromagnetic wave
D. The speed of light
3. (#40 from the text) Place the following types of radiation in order of increasing wavelength and increasing energy per
photon:
Gamma Rays // Radio Waves // Microwaves // Visible Light
Increasing Wavelength:
Gamma Rays <<Visible Light << Microwaves << Radio Waves
Increasing Energy:
Radio Waves << Microwaves << Visible Light << Gamma Rays
4. If an X-ray machine emits E.M.R with a wavelength of 1.00 x 10 -10 meters, what is the frequency?
Use: c = λ v…so:
v=c/λ
v = (3.00x108)/(1.00x10-10)
v = 3.00x1018 s-1
Answer = 3.00 x 1018 Hz
5. A police officer uses a radar gun to catch Heisenberg speeding. The gun operates at a frequency of 22.235 x 10 9 Hz. Find
the wavelength in nanometers of this radiation. (Bonus: What did Heisenberg say to the police officer when the police
officer asked if he knew how fast he was going?)
Use: c = λ v…so:
λ = c/v
λ = (3.00x108)/(22.235x109)
λ = .01349m = 1.349x10-2m = 1.349x107nm
Heisenberg said: I have no idea how fast I was going, but I knew exactly where I was!!
(the officer was not amused by the smartass comment and doubled the fine)
((this is about as good as chemistry humor gets))
Answer = 1.3483 x 107 nm
6. Describe the two types of interference:
1) Constructive. The waves are in phase and the amplitudes sum together.
2) Destructive. The waves have opposite phase and the amplitudes cancel.
(does matter behave like this? NO!!)
Chem 31 SI 11/01/09
Chapter 7 Worksheet
7. During constructive interference:
A. The amplitudes are summed together
B. The wavelengths are summed together
C. The light waves bend around an object
D. The light waves disperse into their component colors
8. What might the problem be if our retina could detect low frequency electromagnetic radiation?
We would experience sensory overload!!!!!
9. What does it mean to be quantized vs. continuous?
Continuous involves gradual change. Quantized involves instantaneous change from one position to the
next, with zero time spent in the intermediate state
10. (for Practice 7.2 – page 290) If a 100 watt light bulb radiates energy at a rate of 100 J/sec, and all of the light emitted
has a wavelength of 525 nm, how many photons are emitted per second?
STEP 1: Find the energy (in Joules) of a single photon @ 525nm
STEP 2: Divide this energy into 100J to find the # of photons emitted in one second
1. Ephoton = hc/λ = (6.626 x 10-34 J ∙ Second)(3.00x108)/(525x10-9m)
Ephoton = 3.786x10-19J
2. (100 J/s) / ( 3.786x10-19J) = 2.641x1020
Answer: 2.64 x 1020 photons
11. What 2 experiments lead to the theory that light does in fact have particle like behavior?
1. Experiments demonstrating the photoelectric effect.
2. Experiments demonstrating the line-spectra of hydrogen.
12. Why does the Bohr model of the Hydrogen atom defy classical physics?
Don’t worry about this question.
13. True or False: The term “orbital” in the quantum mechanical model of the atom is essentially the same thing as
“orbits” in the Bohr model of the atom.
Orbitals in quantum mechanics represent a plot of the probability of an electrons existence and have no
resemblance to a “circular” orbit like that of planetary motion.
14. (#27-29 from text) What are the possible values of n, l, and ml? What does each of these determine?
n = the principal quantum number = specifies the size and energy of the orbital
n can equal any positive integer (1, 2, 3, 4, etc…)
l = the angular momentum quantum number = specifies the shape of the orbital
l is all whole numbers between zero and n-1…so if n = 3, l = 0,1, and 2
l=0=s
l=1=p
l=2=d
l=3=f
ml = magnetic quantum number = specifies the orientation of the orbital
ml = 0 and +/- l…..so if l = 3, ml = -3, -2, -1, 0, 1, 2, and 3
(this really means is that there are seven possible orientations of the f orbital…so don’t get too
hung up on -3, -2, etc…they are abstract…the total # of ml values is what matters)
Chem 31 SI 11/01/09
Chapter 7 Worksheet
15. (#67d from text) Calculate the wavelength of the light emitted when an electron in a hydrogen atom makes the
transition from n = 5 to n = 2 and indicate the region of the electromagnetic spectrum where the light is found:
∆Eatom = Efinal – Einitial = hv = hc/λ = -2.18 x 10-18 J
c = 2.99792458 x 108 meters/second
h = planck’s constant = 6.626 x 10-34 J ∙ Second
Use the bold equation in the box above.
Note that the equation reads: (1/nfinal2 – 1/ninitial2).
Nfinal = 2, and ninitial = 5. We need to solve for λ:
hc/ λ = .-2.18x10-18 [(1/22)-(1/52)]
BUT: BE CAREFUL… hc/ λ =∆Eatom. The atom loses / releases energy since the electron relaxes,
so we need to make the left side of the equation NEGATIVE!!!
- [(6.626 x 10-34 J ∙ Second)(3.00x108)]/ λ = -2.18x10-18 [(1/22)-(1/52)]
- [(6.626 x 10-34 J ∙ Second)(3.00x108)]/ λ = (-2.18x10-18)(.21)
- [(6.626 x 10-34 J ∙ Second)(3.00x108)]/ λ = (-2.18x10-18)(.21)
- (1.9878x10-25)/( λ ) = -4.578x10-19
λ = 4.34x10-7m
Answer = 4.34 x 10-7 m or 434 nm. = Visible Light (Blueish)
16. (#70 from text) An electron in a hydrogen atom relaxes to the n = 4 level, emitting light of 114 THz. What is the value
of n for the level in which the electron originated?
1. Use the equation in the box above. First let’s solve for the left half of the equation hv:
(6.626 x 10-34 J ∙ Second)(114 x 1012 s-1) = 7.55364 x 10-20 J
2. So the change in energy ∆E = 7.55 x 10-20 J. Now, is this number positive or negative? Well,
the atom loses / releases energy since the electron relaxes, so ∆E = - 7.55364 x 10-20 J.
3. Now solve for ninitial. (nfinal = 4) Note that the equation reads: (1/nfinal2 – 1/ninitial2) so:
- 7.55364 x 10-20 J = (-2.18x10-18)(1/42 – 1/ninitial2)
.03464 = (1/16 – 1/ninitial2)
.03464 + (1/ninitial2)= 1/16
(1/ninitial2) = (1/16) – (.03464)
(1/ninitial2) = .02785
1 = (.02785)(ninitial2)
35.9 = ninitial2
5.992 = ninitial
This value is very close to a whole number, so:
Ninitial = 6 !!
Note: the trick was to keep track of the signs. During the test, if you are solving for n and do not
get a solution that is close to a whole number, check your math closely (do not round between each
step), and try flipping the sign on the left side of the equation!
Answer: n = 6
17 (BONUS) You are at a baseball game and a person sitting next to you is trying to impress their date by telling them that
the baseball has vibrations, and that its position in space cannot totally be determined by the batter. You know that such
concepts only really apply to infinitesimally small particles such as the electron. Using the deBroglie relationship and the
Heisenberg uncertainty principle, compare an electron in motion vs. a baseball in motion to mathematically prove your
point (assume the mass of the baseball is 143g and it travels at 95 mph , the mass of an electron is 2.74 x 10-31 kg, assume
an electron is traveling at a speed of 2.65 m/s).
Refer to the deBroglie equation and the Heisenberg equation. The point is that since the mass is so
large the uncertainty in speed and position are infinitesimally small, and that the wavelength
(vibrations) of the baseball are infinitesimally small compared to its diameter.
Chem 31 SI 11/01/09
Chapter 7 Worksheet
18. Which of the following accurately describes l (the angular momentum quantum number)?
A. The value of l can be both positive and negative
B. The value of l specifies the orientation of the orbital
C. The value of l specifies the energy level of the orbital.
D. The number of possible l values equals the value of n.
19. List all of the possible values of l (angular momentum) and m (magnetic) quantum numbers for the following values of
n (the principal quantum number):
A) n = 1
1. n = 1, l = 0, ml = 0
B) n = 2
1. n = 2, l = 0, ml = 0
2. n = 2, l = 1, ml = -1
3. n = 2, l = 1, ml = 0
4. n = 2, l = 1, ml =1
C) n = 3
1. n = 3, l = 0, ml = 0
2. n = 3, l = 1, ml = -1
3. n = 3, l = 1, ml = 0
4. n = 3, l = 1, ml =1
5. n = 3, l = 2, ml = -2
6. n = 3, l = 2, ml = -1
7. n = 3, l = 2, ml = 0
8. n = 3, l = 2, ml =1
9. n = 3, l = 2, ml =2
D) n = 4
1. n = 4, l = 0, ml = 0
2. n = 4, l = 1, ml = -1
3. n = 4, l = 1, ml = 0
4. n = 4, l = 1, ml = 1
5. n = 4, l = 2, ml = -2
6. n = 4, l = 2, ml = -1
7. n = 4, l = 2, ml = 0
8. n = 4, l = 2, ml = 1
9. n = 4, l = 2, ml = 2
10. n = 4, l = 3, ml = -3
11. n = 4, l = 3, ml = -2
12. n = 4, l = 3, ml = -1
13. n = 4, l = 3, ml = 0
14. n = 4, l = 3, ml = 1
15. n = 4, l = 3, ml = 2
16. n = 4, l = 3, ml = 3
20. Based on the above question, the total number of orbitals possible for a given n-value is equal to:
A. n
B. n2
C. n-1
D. 2n + 1
21. Supply the missing quantum numbers and sublevel names:
n
l
ml
Name
4
1
0
4p
2
1
0
2p
3
2
-2
3d
2
0
0
2s
22. Indicate why the following sets of quantum numbers are wrong:
A) n =1 , l = 1 , ml = 0 , Name = 1p
l should = 0 and name should be 1s.
B) n = 4 , l = 3 , ml = +3 , name = 4d
l = 3 = f (not d). So the name should be 4f
C) n = 3 , l = 1 , ml = -2 , Name = 3p
If l = 1, ml cannot = -2….. ml can only be -1, 0, or +1
D) Name = 2d
No 2d orbital exists….the quantum numbers would be n = 2, l = 2, which is not allowed.
Chem 31 SI 11/01/09
Chapter 7 Worksheet
23. (# 78 from text) The energy required to ionize sodium is 496 kJ / mol. What minimum frequency
of light is required to ionize sodium? (refer back to the photoelectric experiment, and refer ahead to ionization
energy)
hv = φ
h = planck’s constant = 6.626 x 10-34 J ∙ Second
v = frequency of light
φ = the energy of a photon when threshold (dislodging of electron) is achieved
The trick is that we want to know how much energy to ionize a
sodium atom. So we need to convert moles to # of atoms using
avogadro’s number. Since h is in terms of joules, we need to convert
kJ to J. Then we can solve for frequency.
1. (496 kJ / mol) * (1mol / 6.022 x 1023) * (1000 J / 1 kJ) = 8.236 x 10-19 Joules per atom = φ
2. v = φ/h
-19
v = (8.236 x 10 Joules /atom) / (6.626 x 10-34 Joules ∙ Second)
v = 1.24 x 1015 cycles per second
Answer = 1.24 x 1015 Hz
24. (#83 from text) The binding energy of electrons in a metal is 193 kJ / mol. Find the threshold
frequency of the metal.
hv = φ
1. Since φ = 193 kJ / mol, convert to Joules per atom or photon:
(193 kJ / mol) * (1mol / 6.022 x 1023) * (1000 J / 1 kJ) = 3.2049 x 10-19 Joules per atom= φ
2. hv = φ
-34
(6.626 x 10 J ∙S)(v)= (3.2049 x 10-19 J/atom)
v = 4.836x1014 s-1 per photon
Answer = 4.84 x 1014 Hz
Chem 31 SI 11/01/09
Chapter 7 Worksheet
Chapter 7 Useful Information
c=λ ∙ ν
c = 2.99792458 x 108 meters/second = the speed of light = the speed of any EM wave in a vacuum
λ = wavelength
ν = frequency (measured in Hz or cycles per second)
λ ultraviolet light = 1 x 10-8 meters
λ infrared light = 1 x 10-4 meters
h = 6.626 x 10-34 Joules ∙ Second = Plancks constant
∆Eatom = E emitted (or absorbed) radiation = hν = Ephoton = hc/λ
DeBroglie relationship between wavelength (λ), mass (m), and velocity (v) of a particle:
λ = h/mv
φ = energy of a photon when threshold is achieved
hv = φ
Calculating the difference between two energy levels in the H atom:
∆E = Efinal – Einitial = hc/λ = -2.18 x 10-18 J
Heisenberg Equation:
∆x + m∆v ≥ h/4π
Visible spectrum = 400 – 750nm