CHEM101, B2
HMWK 02 Answers
2008 01 21
EMR - Bohr
HT
01.)
0.87 µm
a.)
b.)
c.)
d.)
The above depicts the wave propagation of a particular EMR.
What is a.) its wavelength, b.) its frequency?
c.) In what region of the EMR spectrum does it occur?
d.) What is its color, if any?
the arrow spans 1.5 wavelengths (crest to crest to trough);
therefore, λ = 0.87 µm ÷1.5 = 0.58 µm = 5.8 x 10–7 m = 580 nm
since we are dealing with light , c = 3.00 x 108 m/s and
3.00 x 108 m/s
ν = c/λ = 5.8 x 10–7 m = 5.17 x 1014 s-1 = 5.17 1014 Hz
this is in the visible part of the spectrum.
using a color wheel, e.g. chart 6 of the CDS, we estimate the color to be yellow/orange , similar
to the sodium flame.
02.) Photoelectric Effect. Light with a wavelength of 400 nm strikes the surface of cesium(Cs) in
a photoelectric cell, and the kinetic energy (KE) of electrons released is 1.54 x 10–19 J.
Calculate
a.) the work function (binding energy, Φ) for Cs.
b.) the longest wavelength of light capable of ejecting e – ’s from the Cs surface.
a.) Acc. to Planck, photons of λ = 400 nm possess an energy of
hc 6.63 x 10-34 Js × 3.00 x 108 m/s
=
= 4.97 x 10-19 J
E = hν =
4.00 x 10-7 m
λ
Part of this energy is needed to release the electron from the surface (Φ) and the rest is
converted into kinetic energy (KE) of the released electron.
therefore, energy of photon (E) = (Φ) + (KE) and (Φ) = (Ε) − (ΚΕ)
Φ = 4.97 x 10-19 J - 1.54 x 10–19 J = 3.43 x 10–19 J
2
b.)
This is the minimum energy required to get release of an electron. Using the Planck equation
again we can find the least energetic radiation ( = longest wavelength ) that corresponds to
this energy.
hc
hc
E = , therefore λ =
Ε
λ
-34
6.63 x 10 Js × 3.00 x 108 m/s
λ=
= 5.80 x 10-7 m = 580 nm
3.43 x 10–19 J
Anything weaker than yellow light (having longer wavelength) will not release electrons from
this surface.
03.) Associate the most appropriate of the following terms: “diffraction, interference, refraction,
dispersion” with statements a.) - d.)
a.) isolation of light rays with different wavelengths dispersion
b.) passage of a wave front through a slit with dimensions similar to the wavelength diffraction
c.) bending of light rays at a water/air surface refraction
d.) interaction of two or more waves interference
04.) How many beryllium, Be, atoms could be lined up within one wavelength of red light?
a.) “Ballpark”: We know that the wavelength of visible light is in the 100’s of nanometers , while
the diameters of atoms are in the 100 of picometers. So, the short answer is : about a 1000.
b.) Detailed calculation: r(Be) = 111 pm (from Pet. p.352);
therefore, diameter = 222 pm = 0.222 nm
λ(red) = 750 nm (from color wheel)
1 atom
# of Be atoms per 750 nm: 750 nm × 0.222 nm = 3380 atoms.
Our initial estimate is not that far off. Purpose of this exercise is to get the message that the
wavelength of visible light is relatively long when compared with the size of atoms.
05.) Can the lines observed in the flame test for lithium (Li) be calculated by the theory developed
in class? Explain qualitatively.
Some of the lines could be, but only those involving the hydrogenic (one electron) species, Li 2+.
Far more complicated theory is needed to determine emission lines resulting from transitions
in the Li and Li+ species.
The dominant red line actually results from a transition in the neutral Li species.
3
06.) In the hydrogen spectrum, do any of the Paschen lines overlap with the Balmer lines?
The lowest energy (longest wavelength) Balmer line is the transition from n=3 to n=2.
The highest energy (shortest wavelength) Paschen line is the transition from n= ∞ to n=3.
1 1
1 1
lowest Balmer: ∆E 3→2 = - RH (n 2 -n 2) = - 2.18 × 10–18 J (22 - 32) = - 3.03 × 10–19 J
f
i
Eph = + 3.03 × 10–19 J
1 1
1 1
highest Paschen: : ∆E ∞→3 = - RH (n 2 -n 2) = - 2.18 × 10–18 J (32 - 2) = - 2.42 × 10–19 J
∞
f
i
–19
Eph = + 2.42 × 10 J
This means that the “weakest” Balmer line is stronger than the “strongest” Paschen line;
and in turn, that the longest Balmer λ is shorter than the shortest Paschen λ .
Therefore, there is no overlap between the two series.
07.) Describe the chemical/physical process that occurs in neon lights.
Glass tubes are filled with neon gas at low pressure (similar to hydrogen gas discharge tubes).
A voltage difference is applied to the cathode and anode in the tube.
If the voltage difference is high enough electrons are released from the cathode surface.
These are able to produce excited neon atoms; i.e., lift electrons to a higher energy level.
As these electron return to a lower energy level, photons (in the form of the typical red neon light)
are released.
08.) Determine an electronic transition for the He+ species that occurs in the visible part of the
EMR spectrum, if any.
For the He+ species we can derive the following formula for its emission lines:
hc
λ= E =
hc
hc
1
6.63 x 10-34 Js × 3.00 x 108 m/s
1
=
×
=
×
2
-18
2
RH Z
2.18 x 10 J x 2
1 1
1 1
1 1
RH (n 2 -n 2) Z2
(n 2 -n 2)
(n 2 -n 2)
f
i
f
i
f
i
= 2.28 x 10-8 m ×
1
1 1
(n 2 -n 2)
f
i
= 22.8 nm ×
1
1 1
(n 2 -n 2)
f
i
4
To get an idea let’s try transitions from ∞ to various nf values.
λ = 22.8 nm ×
1
1
= 22.8 nm × nf2
1 1 = 22.8 nm × 1
(n 2 - 2)
(n 2 -0)
f ∞
f
The value for nf2 should be between ~ 20 and 40 to get into the range of 400 - 800 nm.
So 25 would be good; corresponding to nf = 5.
To double check: λ∞→5 = 22.8 nm × 52 = 570 nm.
The transition from n=∞ to n=5 gives a line at 570 nm , which is yellowish.
Of course, there are other ways to solve this; the above transition is certainly not the only one
that provides visible light from the He+ species.
09.) a.) For lithium what ionization energy can be calculated by formulas developed in class?
We can deal only with hydrogenic atoms, here only with transitions in the Li2+ species.
So we can calculate the transition of the third electron from n=1 to n=∞
(after the first 2 electrons have already been removed.)
b.) Write a “chemical equation” for this process.
Li2+(g) → Li3+(g) + e– (I have added the subscript g to indicate that this occurs in the gas state)
c.) What is the value for this ionization energy?
1 1
1 1
∆E 1→∞ = - RH (n 2 -n 2) Z2 = - 2.18 × 10–18 J ( 2 - 12) 32 = - 2.18 × 10–18 J (0 - 1) 32
∞
f
i
= + 1.96 x 10-17 J
( I emphasize that this is a positive system energy change; the transition might come from the
impact of a photon, but not necessarily so)
10.) In the emission spectrum of Li2+ species what is a.) the shortest, b.) the longest wavelength
where the final energy level nf =3.
1 1
1 1
a.) shortest : ∆E ∞→3 = - RH (n 2 -n 2) Z2 = - 2.18 × 10–18 J (32 - 2) 32 = - 2.18 × 10–18 J
∞
f
i
Eph = + 2.18 × 10–18 J
hc 6.63 x 10-34 Js × 3.00 x 108 m/s
λ=E =
= 9.12 x 10-8 m = 91.2 nm
2.18 x 10-18 J
ph
5
1 1
1 1
b.) longest : ∆E 4→3 = - RH (n 2 -n 2) Z2 = - 2.18 × 10–18 J (32 - 42) 32 = - 9.54 × 10–19 J
f
i
Eph = + 9.54 × 10–19 J
hc 6.63 x 10-34 Js × 3.00 x 108 m/s
λ=E =
= 2.08 x 10-7 m = 208 nm
–19
J
9.54
×
10
ph
11.)
What is the total energy released when 1 mol of Li2+ ions undergoes the transition described
in 10.a) ?
The energies calculated above relate to one atomic event; if we want to know the energy
for 1 mol of transitions we simply multiply by Avogadro’s number:
2.18 x 10–18 J 6.02 x 1023 photons
= 1.31x 106 J/mol = 1.31x 103 kJ/mol
1 photon ×
1 mol
12.) Sketch the lines of the Li2+ emission spectrum where nf =3; label each line indicating the
transition(ni → nf ); also place lines that correspond to the transition n=5 to n=4 (label “A”)
and to the ultimate ionization energy (label “B”).
∞→1
∞→3
.......... 6→3
5→3
4→3
B
5→4
A
wavelength
13.) Generally, what is the system energy change during absorption?
Absorption in this context refers to disappearance of light or disappearance of a photon;
“system” means “nucleus + electron(s)”.
During absorption an electron moves further away from the nucleus; i.e., the system energy
increases.
The disappearance of a photon can be considered a negative energy change for the photon,
thereby satisfying ∆Esys + ∆Eph = 0, the Law of Conservation of Energy.
(system energy ↑, photon energy ↓)