Phys 1251
Solutions: Group Work 13 – Spectroscopy
Problem 3 (Pg. 21)
The work function of a metal surface is 2.85 eV. Find the maximum kinetic energy of electrons
ejected from this surface by light of wavelength
(a) 376 nm.
The kinetic energy of the photoelectrons is given by the expression
Kmax = hf − Φ,
where hf is the energy of the bombarding photons and Φ is the work function of the metal. Since we know
the wavelength of the photons, we can write
Kmax =
Kmax =
hc
−Φ
λ
hc
− 2.85 eV
376 nm
Kmax = 0.45 eV
(b) 452 nm.
Again, we can use the relation
Kmax =
hc
− Φ.
λ
However, when we apply this equation we find that
Kmax = −0.11 eV
Electrons cannot be ejected from the metal if their kinetic energy is less than zero, because then they will not
have enough energy to overcome the attractive potential of the positive atomic cores on the metal surface.
Therefore we conclude that
Kmax = 0
and there are no electrons ejected.
(c) What is the cutoff wavelength?
The cutoff wavelength is the wavelength above which no electrons will be ejected from the metal because
the incoming photons will not have enough energy to overcome the work function of the metal. We can find
this cutoff wavelength by setting the maximum kinetic energy of the photoelectrons just equal to zero.
0=
hc
−Φ
λ
λ=
hc
Φ
λ = 435 nm
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Phys 1251
Solutions: Group Work 13 – Spectroscopy
Problem 6 (Pg. 21)
An electron is confined to a rigid box that is 1.20 nm long.
(a) If the electron is in its third excited state (n = 4), can you find its de Broglie wavelength
without first finding the energy? If so, what is the wavelength?
Just as we can find the wavelengths of higher harmonics of an oscillating guitar string, we can find the
wavelength of the higher excited states of a particle in a box if we know the size of the box. Since the
fundamental wavelength (longest wavelength) of the particle in a box is λ1 = 2L (because this configuration
gives one antinode; see the left side of the figure below), the wavelength in any state n is given by
λn =
2L
n
Therefore the wavelength of the third excited state is
λ4 =
2L
2 · 1.20 nm
=
4
4
λ4 = 0.6 nm = 6 Å
Remember that since the ground state is the n = 1 state, the third excited state is n = 4.
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Phys 1251
Solutions: Group Work 13 – Spectroscopy
(b) If the electron is in its third excited state (n = 4), what are all the possible wavelengths
of light that the electron can emit as it makes its way back down to the ground state?
To get back to the ground state electron doesn’t simply have to jump directly from the n = 4 state to the
n = 1 state. It can take any one of a number of paths from the n = 4 state to the n = 1 state, stopping at
either or both of the states in between. As illustrated in the right side of the figure above, this means that
there are six different wavelengths that can be emitted, corresponding to the six possible transitions. The
energy of each state is given by
h2 n2
En =
8mL2
and the energy of an emitted photon is equal to the energy difference between two states,
Ephoton = ∆E =
h2 2
2
n
−
n
1 .
8mL2 2
Additionally, the energy of a photon is related to its wavelength by
Ephoton =
hc
,
λ
so by relating the photon wavelength to the energy difference between allowed states we find
λ=
8mcL2
.
h n22 − n21
Therefore our six emitted wavelengths (as labeled on the graph above) are
λ1 = 317 nm
λ2 = 396 nm
λ3 = 680 nm
λ4 = 590 nm
λ5 = 950 nm
λ6 = 1580 nm
Problem 8 (Pg. 21)
An electron is accelerated from rest through a potential difference so that it has a de Broglie
wavelength of 400 pm.
(a) What potential difference is needed?
If the electron has a de Broglie wavelength of 400 picometers, then it has a momentum of
p=
h
h
=
= 1.66 × 10−24 kg · m/s
λdeBroglie
400 × 10−12 m
We can relate the momentum to the electrons kinetic energy:
K=
1
p2
me v 2 =
2
2me
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Phys 1251
Solutions: Group Work 13 – Spectroscopy
If the electron is being accelerated from rest, then the potential difference it moves through is equal to its
initial potential energy. Therefore all its initial potential energy is converted to kinetic energy and
K = qV
p2
= eV
2me
V =
h2
2me eλ2deBroglie
V = 9.4 V
(b) What would the potential difference need to be to accelerate a proton until it has that
same wavelength?
We can take the same expression that we derived above in part (a) but substitute the proton mass for the
electron mass:
V =
h2
2mp eλ2deBroglie
V = 5.12 mV
Since the proton’s mass is approximately 2000 times greater than the electon’s mass, the potential needed
to accelerate the proton is 2000 times smaller than that for the electron.
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