Light, Photon Energies, and Atomic Spectra λ ν λ ν

Electromagnetic Radiation
Light, Photon Energies,
and Atomic Spectra
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Electromagnetic radiation (radiant energy) is
characterized by its:
z
z
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wavelength (color): λ (Greek letter lambda)
frequency (energy): ν (Greek letter nu)
Electromagnetic Radiation
Waves
They are related by the equation:
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ν=
z
Wavelength = distance between
successive “crests”
| Frequency = the # of crests passing
a given point per second
c
λ
where c = 3.00 x 108 m/s
(the speed of light in a vacuum)
Example: The frequency of violet light is
7.31 x 1014 Hz, and that of red light is 4.57
x 1014 Hz. Calculate the wavelength of
each color.
1 Hz = sec-1 or 1/sec
Violet:
7.31× 1014 Hz =
3.00 × 10
ν=
8 m
s
λ
3.00 ×108 ms
λ=
7.31×1014 1s
λ = 4.10 ×10 −7 m
c
λ
Example: The frequency of violet light is
7.31 x 1014 Hz, and that of red light is 4.57
x 1014 Hz. Calculate the wavelength of
each color.
1 Hz = sec-1 or 1/sec
Red:
4.57 × 1014 Hz =
3.00 × 10
ν=
8 m
s
c
λ
λ
3.00 ×108 ms
λ=
4.57 ×1014 1s
λ = 6.56 × 10 −7 m
1
Light
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Photon Energy
When sunlight or white light is passed
through a prism, it gives the continuous
spectrum observed in a rainbow.
We can describe light as composed of
particles, or PHOTONS.
Each photon of light has a particular amount
of energy (a quantum).
The amt. of energy possessed by a photon
depends on the color of the light.
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The energy of a photon is given by
this equation:
E = hν
z
Example: Calculate the energy, in joules,
of an individual photon of violet and red
light.
ν violet = 7.31×1014 Hz
Violet:
ν red = 4.57 ×1014 Hz
E = (6.6262 × 10 J ⋅ s )(4.57 × 1014 Hz )
Red:
What does this have to do with
electron arrangement in atoms?
When all electrons are in the lowest
possible energy levels, an atom is
said to be in its GROUND STATE.
| When an atom absorbs energy so that
its electrons are “boosted” to higher
energy levels, the atom is said to be in
an EXCITED STATE.
E = (6.6262 × 10 -34 J ⋅ s )(4.57 × 1014 1s )
E = 3.03 ×10 −19 J
E = 4.84 × 10 −19 J
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E = hν
-34
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E = (6.6262 × 10 -34 J ⋅ s )(7.31× 1014 1s )
h = 6.6262 x 10-34 J•s
ν = frequency (Hz)
Example: Calculate the energy, in joules,
of an individual photon of violet and red
light.
E = hν
E = (6.6262 ×10 J ⋅ s )(7.31×10 Hz )
-34
where
Bright Line Emission Spectrum
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The light emitted by
an element when its
electrons return to a
lower energy state
can be viewed as a
bright line emission
spectrum.
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Absorption Spectrum
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Light
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The light absorbed by an element when
white light is passed through a sample is
illustrated by the absorption spectrum.
Note: The wavelengths of light that are
absorbed by the gas show up as black lines,
and are equal to the wavelengths of light
given off in the emission spectrum.
Why?
z
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Absence of color appears black. Only
reflected colors of light are visible.
Example: A green line of wavelength 486
nm is observed in the emission spectrum
of hydrogen. Calculate the energy of one
photon of this green light.
λgreen = 486 ×10 m
−9
Green:
hc
E =
λ
E =
ν=
c
Example: The green light associated with
the aurora borealis is emitted by excited
(high-energy) oxygen atoms at 557.7 nm.
What is the frequency of this light?
E = hν
λ
Electronic energy is quantized (only certain
values of electron energy are possible).
When an electron moves from a lower
energy level to a higher energy level in an
atom, energy of a characteristic frequency
(wavelength) is absorbed.
When an electron falls from a higher energy
level back to the lower energy level, then
radiation of the same frequency (wavelength)
is emitted.
The bright-line emission spectrum is unique
to each element, just like a fingerprint is
unique to each person.
λ = 557.7 ×10−9 m
Green:
(6.6262 × 10 -34 J ⋅ s )(3.00 × 10 8
-9
486 × 10 m
m
s)
3.00 ×108 ms
ν=
557.7 ×10-9 m
ν=
c
λ
ν = 5.38 × 1014 Hz
E = 4.09 ×10 −19 J
3