17/09/2010
Clicker questions and Lecture questions
(Section 6.1-6.4) as of Friday Sept 17
Which wave is of higher energy?
Example:
What is the wavelength (in nanometers) of uv
light with  = 5.5 x 1015 s-1?
c = λ ν, rearrange to give λ = c/ν
1nm = 1 x 10-9 m

3.00 x108 m / s
5.5 x1015 / s
λ = 5.4 x 10-8 m = 54 x 10-9 m = 54 nm
Units OK for wavelength (distance),
is wavelength correct for uv light?
Chem 101
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17/09/2010
Human skin is not penetrated by light but is penetrated by
X-rays. Which travels faster, X-rays or visible light?
1. Both visible light and X rays travel at the same
speed, about 186,000 miles per second.
2. Additional energy information is needed to
compare the speeds.
3. X rays travel faster.
4. Visible light travels faster.
What is the difference between visible light and
electromagnetic radiation?
A.
B.
C.
D.
E.
Electromagnetic radiation contains both an oscillating electric field
and an oscillating magnetic field; visible light only contains one of the
two.
Visible light is only one form of electromagnetic radiation; all visible
light is electromagnetic radiation, but not all electromagnetic
radiation is visible light.
Electromagnetic radiation is only found at higher energies than
visible light.
Electromagnetic radiation is only one form of visible light; all
electromagnetic radiation is visible light, but not all visible light is
characterized as electromagnetic radiation.
Visible light can be seen with the naked eye, but electromagnetic
radiation is invisible.
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What is the difference between visible light and
electromagnetic radiation?
B.
Visible light is only one form of electromagnetic radiation; all
visible light is electromagnetic radiation, but not all
electromagnetic radiation is visible light.
The temperature of stars may be gauged by their colour, e.g.
blue-white stars are hotter than red stars. How is this
observation consistent with Planck’s assumption?
A.
B.
C.
D.
E.
Planck proposed that matter is allowed to emit energy across a
continuous spectrum analogous to the spectrum of visible light.
Planck proposed that all types of matter would emit the same frequencies
of visible light when at the same temperatures.
As temperature increases, so does the average energy of the emitted
radiation. Blue-white light is at the longer wavelength (lower energy) end
of the visible spectrum while red light is at the shorter wavelength (higher
energy) end of the visible spectrum.
As temperature increases, so does the average energy of the emitted
radiation. Blue-white light is at the shorter wavelength (higher energy)
end of the visible spectrum while red light is at the longer wavelength
(lower energy) end of the visible spectrum.
Search me.
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17/09/2010
The temperature of stars may be gauged by their colour, e.g.
blue-white stars are hotter than red stars. How is this
observation consistent with Planck’s assumption?
D.
As temperature increases, so does the average energy of the
emitted radiation. Blue-white light is at the shorter wavelength
(higher energy) end of the visible spectrum while red light is at the
longer wavelength (lower energy) end of the visible spectrum.
Suppose that yellow visible light can be used to
eject electrons from a certain metal surface. What
would happen if ultraviolet light was used instead?
A.
B.
C.
D.
No electrons would be ejected.
Electrons would be ejected, and they would have the
same kinetic energy as those ejected by yellow light.
Electrons would be ejected, and they would have
greater kinetic energy than those ejected by yellow
light.
Electrons would be ejected, and they would have lower
kinetic energy than those ejected by yellow light.
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17/09/2010
Suppose that yellow visible light can be used to
eject electrons from a certain metal surface. What
would happen if ultraviolet light was used instead?
C) Electrons would be ejected, and they would have
greater kinetic energy than those ejected by yellow
light.
Photoelectric effect
In separate experiments, light of
three different wavelengths –
325 nm, 455 nm and 632 nm
– shone on a metal surface
and the following
observations were made.
Which one of the statements
below is correct?
Wavelength A: no photoelectrons
were observed
Wavelength B: photoelectrons with
a kinetic energy of 155 kJ/mol
were observed
Wavelength C: Photoelectrons
with a kinetic energy of 51
kJ/mol were observed
 A) Observation A
corresponds to 455 nm
 B) Observation B
corresponds to 325 nm
 C) Observation C
corresponds to 632 nm
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17/09/2010
Photoelectric effect
In separate experiments, light of
three different wavelengths –
325 nm, 455 nm and 632 nm
– shone on a metal surface
and the following
observations were made.
Which one of the statements
below is correct?
Wavelength A: no photoelectrons
were observed
Wavelength B: photoelectrons with
a kinetic energy of 155 kJ/mol
were observed
Wavelength C: Photoelectrons
with a kinetic energy of 51
kJ/mol were observed
 A) Observation A
corresponds to 455 nm
 B) Observation B
corresponds to 325 nm
 C) Observation C
corresponds to 632 nm
Example:
MRI body scanners operate with 400 MHz radio frequency
energy. How much energy does this correspond to in
kilojoules/mol?
Energy = h x (frequency) or E = h ν
hertz (Hz) is sec-1 and
M means mega (106)
Note: the equation gives the energy of one photon, problem
asks for energy of 1 mol of photons.
Will need to use Avogadro’s number to calculate energy
for one mol.
E = (6.63 x 10-34 J s) (400 x 106 s-1) = 265 x 10-27 J
= (265 x 10-27 J )(10-3kJ/J) = 265 x 10-30 kJ = 2.65 x 10-28 kJ per photon
Energy per mol = (E/photon) (photons/mol )
= (2.65 x 10-28kJ/photon)(6.023 x 1023photons/mol) = 1.60 x 10-4 kJ/mol
Units are correct, answer seems reasonable
Chem 101
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17/09/2010
As the electron in a hydrogen atom jumps from the n = 3
orbit to the n = 7 orbit, does it absorb energy or emit
energy?
A.
B.
C.
D.
E.
It neither emits nor absorbs energy.
It both emits and absorbs energy simultaneously.
It emits energy.
It absorbs energy.
Hmm, a 50:50 proposition… Heads!
As the electron in a hydrogen atom jumps from the n = 3
orbit to the n = 7 orbit, does it absorb energy or emit
energy?
A.
B.
C.
D.
E.
It neither emits nor absorbs energy.
It both emits and absorbs energy simultaneously.
It emits energy.
It absorbs energy.
Hmm, a 50:50 proposition… Heads!
absorption
excited state
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17/09/2010
Speculate as to why the Balmer series (electrons falling from
higher orbitals to n=2) was known earlier than the Lyman
series (electrons falling from higher orbitals to n=1) despite
the fact that falling to the ground state (n=1) is more likely.
A.
B.
C.
D.
E.
Under Bohr’s model, the ground state is not allowed as it would
cause nuclear collapse.
The Lyman series is an absorption spectrum while the Balmer series
is an emission spectrum and it is easier to detect emission.
The Balmer series occurs within the visible region of the spectra and
it was easier to detect, as it is simpler to detect visible light.
The Lyman series emits photons in the hard UV region and everyone
trying to detect those photons got cancers and perished before their
results were published.
I’m too cautious to speculate.
Speculate as to why the Balmer series (electrons falling from
higher orbitals to n=2) was known earlier than the Lyman
series (electrons falling from higher orbitals to n=1) despite
the fact that falling to the ground state (n=1) is more likely.
C.
The Balmer series occurs within the visible region of the spectra and
it was easier to detect, as it is simpler to detect visible light.
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17/09/2010
A baseball pitcher throws a fastball at 150 km/h. Does that
moving baseball generate matter waves? If so, can we
observe them?
A.
B.
C.
D.
E.
No matter waves are produced.
No, because the mass of the baseball is too large.
Yes; but too small to allow any way of observing them.
Yes; and they can be observed.
Let me ask YOU; what is the sound of one hand clapping?
A baseball pitcher throws a fastball at 150 km/h. Does that
moving baseball generate matter waves? If so, can we
observe them?
C.
Yes; but too small to allow any way of observing them.
λ = h/mv
= (6.63 x 10-34 J s)/{(150 g)(150 km/hr)}
λ = {6.63 x 10-34 (kg m 2 /s2) s}/{(0.150 kg)(150x103m/3600s)}
λ = 1.08x10-34m
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Why do we ignore wave properties for most objects?
Example:
What is the deBroglie wavelength (in meters) of a mosquito
weighing 1.55 mg and traveling at 156 km/hr?
1 km/hr = 103m/3600s
λ = h/mv
Units are a problem: h is in J s and the
denominator has mass times m/s
Data Sheet : 1 J = 1 kg m2/s2
Note that the
wavelength is too
short to measure
or matter!
1mg = 10-3 g
λ = (6.63 x 10-34 J s)/{(1.55 mg)(156 km/hr)}
λ = {6.63 x 10-34 (kg m 2 /s2) s}/{(1.55 x 10-3 g)(156x103m/3600s)}
λ = 9.87 x 10-33 kg m/g = (9.87 x 10-33m kg/g) (103g/kg) = 9.87x10-30m

Units are distance, which is correct for wavelength
Chem 101
19
Example:
The electron’s size is 1x10-10m, what is uncertainty in velocity?
(Δx) (Δmv) ≥ h/4π
Units require 1 J = 1 kg m2/s2
Δv ≥ 6.63x10-34 kg m 2/s2) s /4π(1x10-10 m)(9.11x10-31 kg)
Assume the
size of the
atom is the
uncertainty in
position
≥ 5.78x105 m 2/m s
Units correct?
Chem 101
No, must cancel metres
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