Photoelectric effect (Einstein)
Chapter 8
Electrons in Atoms
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1050.php
Light can be treated as particles called
photons. The energy of a photon
depends on the light’s frequency when
treated as a wave (see slide 30), while a
greater intensity of light implies a
greater number of photons.
The slope of
this graph is
Planck’s
constant!
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Energy of a photon is quantized
Explaining the photoelectric effect
When a photon hits an bound electron in
an atom, the electron can absorb the
photon energy.
If the photon energy is greater than the
work function (the amount of energy
required to “just” unbind the electron from
the atom), then the remaining energy goes
into determining the kinetic energy of the
now unbound electron.
E = hν
where
h = 6.62607 x 10-34 J s
Planck’s constant
ν is frequency
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Explaining the photoelectric effect
Explaining the photoelectric effect
To measure the kinetic energy of the
electron, we set up a stopping voltage Vs
between two metal plates. The voltage is
adjusted until the electron stops moving,
which happens when
Ek = ½ me
v2 =
Ek = ½ mev2 = eVs
Here me is the electron rest mass of
9.109 x 10-34 kg
and e is the electron charge
eVs
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1.602 x 10-19 C
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Work function and threshold frequency
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Work function and threshold frequency
We saw there was a threshold ν0 in
the photoelectric effect which defined
the minimum frequency (and
therefore minimum energy of a
photon since E = hν) required to eject
the electron. Any energy above the
threshold is what is measured by the
stopping voltage.
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The work function can be expressed
Ew = hν0
where ν0 is the minimum frequency
of light required to free an electron,
which depends on the metal.
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Work function and threshold frequency
Work function and threshold frequency
By energy conservation, the energy of the
photon must be used to overcome the
work function, and provide the kinetic
energy of the electron, so
Therefore
Ek = hν - Ew
Ew + Ek = hν
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Problem
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Problem answer
The minimum energy (the work function)
required to cause the photoelectric effect
in potassium metal is 3.69 x 10-19 J. Will
photoelectrons be produced when blue
light of wavelength 400 nm is shone on the
metal? If they are ejected, what is the
velocity of the electrons?
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Electrons will be ejected by 400 nm light,
with a velocity of 1.68 x 107 m s-1. This is
about 5.6% of the speed of light.
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3
Photochemistry
Problem
Photons of light can provide enough
energy to break chemical bonds by
changing the distribution of electrons in the
molecule. We can treat hν as a reactant!
O2 + hν ↓ 2 O
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Problem answer
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The Bohr atom
E = 3.056 x 10-19 J photon-1
ν = 4.612 x 1014 s-1 and λ = 650.4 nm
Absorption of red light
and E = 4.414 x 10-19 J photon-1
ν = 6.662 x 1014 s-1 and λ = 450.3 nm
Absorption of blue light
When you absorb red and blue, the color of
visible light that is reflected (and seen!) is around
a wavelength of 550 nm, which is green!
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Chlorophyll absorbs light energies of 3.056
x 10-19 J photon-1 and 4.414 x 10-19 J
photon-1. To what color, frequency and
wavelength do these absorptions
correspond, and can you use these results
to explain why chlorophyll appears green?
Often our vision of the
electrons in an atom
is one of the electrons
“orbiting” the nucleus
like the planets orbit
around the Sun. This
would be classical
mechanics behaviour.
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The Bohr atom
The Bohr atom
The problem with this
image is that an orbiting
electron is always
accelerating – and
accelerating charges
give off light (and
therefore lose energy!)–
the electron should lose
energy and “death spiral”
into the nucleus!
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Niels Bohr proposed
a different model
with three main
properties:
First, the orbit of an
electron is circular
like in classical
physics.
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The Bohr atom
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The Bohr atom
Second, an electron is
only allowed to have of
fixed set of orbits called
stationary states, which
depend on the angular
momentum of the
electron, which depends
π, where n is the
on nh/2π
principal quantum
number, and can only
be a non-zero integer, so
n = 1, 2, 3, #
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Third, an electron can
only move from one
stationary state to
another. This transition
requires the absorption
or emission of a photon
with an energy
matching the difference
of the energy of the
electron in the two
stationary states.
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The Bohr hydrogen atom
Energy level diagram
Radius of an allowed orbit is rn = n2a0
a0 is the Bohr radius where a0 = 53 pm = 0.53 Å
See slide 24!
If we treat the infinitely separated nucleus and
electron as zero in energy, the energy of each
orbit is
=
Second excited state
n=3
First excited state
n=2
−
where RH (the Rydberg
constant) is
2.179 x 10-18 J
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Ground state n = 1
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Problem
Electronic transitions in a hydrogen atom
The photon absorbed or emitted must
have an energy that exactly matches the
energy difference between the two
electronic states
The energy of an electron in a hydrogen
atom is -4.45 x 10-20 J. What energy level
does the electron occupy?
∆E=hν
1
1
=R H
−
ni 2 nf 2
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Problem answer
Problem
n=7
Determine the wavelength of light
absorbed in an electron transition from n =
2 to n = 4 in a hydrogen atom.
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Problem answer
n=5
n=4
The line spectra we saw
earlier are emission
spectra, where light is
given off.
Now we’ve seen that light
can also be absorbed,
which means we can have
absorption spectra
where specific
wavelengths are absorbed
(but usually only from the
ground state).
n=3
The Balmer series represents all
the transitions to or from the n = 2
state!
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Absorption spectra
λ = 486.5 nm
n=6
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Helium absorption spectrum
Helium was actually
discovered from the
absorption spectrum of
the Sun in 1868.
Norman Lockyer
predicted the existence
of helium as an
element 27 years
before it was isolated
on Earth!
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