When treated with nitric acid, lead(IV) oxide reacts according to the following unbalanced
equation:
PbO2 (s) + HNO3 (aq) −→ Pb(NO3 )2 (aq) + H2 O(l) + O2 (g)
The volume of evolved gas can be measured by performing the reaction in an inverted burette.
If 13.4 grams of PbO2 (s) are added to 20.0 mL of 7.91 M HNO3 (aq) at 1.00 atm and 25 ◦ C,
what volume of O2 (g) will be produced?
First we balance the equation. Because of the ratio of Pb to NO3 in the products, the
ratio of PbO2 to HNO3 is 1:2. This leaves 2 protons on the left hand side, so we get 1 water
molecule. Balancing for O2 we get a coefficient of 1/2. Doubling all the coefficients gets
them to whole numbers, we get:
2 PbO2 (s) + 4 HNO3 (aq) −→ 2 Pb(NO3 )2 (aq) + 2 H2 O(l) + O2 (g)
Now we have to determine the limiting reagent.
M MPbO2 = 207.2 + 2(16.0) = 239.2
g
mol
1 mol
= 0.0560 mol
239.2 g
mol
1L
= 7.91
× 20 mL ×
= 0.158 mol
L
1000 mL
nPbO2 = 13.4 g ×
nHNO3
1
2
× 0.560 mol <
1
4
× 0.158 mol , so PbO2 is the limiting reagent.
Now we determine the number of moles of O2 produced.
nO2 = 13.4 g PbO2 ×
1 mol PbO2
1 mol O2
×
= 0.0280 mol O2
239.2 g PbO2 2 mol PbO2
Finally, using the ideal gas law, we determine the volume of O2
P V = nRT
nRT
V =
P
L atm
0.0280 mol × 0.0821 mol
× 298K
K
VO2 =
1.00 atm
VO2 = 685 mL
Consider a violet diode laser emitting a power of 20 mW at a wavelength of 405 nm (these
are the lasers used in Blu-Ray applications). The beam is directed at a clean surface of Li,
which has a work function of φ = 2.93 eV.
Assuming that 1.00% of the indicent photons contribute to the photoelectric effect (the other
99% get reflected or act in some other processes), calculate (a) How many electrons are being
ejected per second, and (b) at what velocity are the electrons ejected?
We start by considering the speed of the ejected electrons. Recall the basic equation of the
photoelectric effect, which is basically a restatement of the law of conservation of energy.
Ek = hν − φ
We know that the kinetic energy is related to the velocity: Ek = 12 mv 2 . Using these relations
and solving for velocity, we have:
1 2
mv = hν − φ
2
r
2(hν − φ)
v=
m
c = νλ
c
ν=
λ
s
v=
− φ)
2( hc
λ
m
Plugging this in m = me = 9.11 × 10−31 kg , h = 6.626 × 10−34 Js , φ = 2.93 eV ×
1.602×10−19 J
= 4.69 × 10−19 J , λ = 405 × 10−9 m , we get
1 eV
v = 2.1 × 105
m
s
To calculate the number of electrons ejected per second, we need to know the number of
photons being emitted per second by the laser. Because each photon caries a set amount of
energy, and we know the total amount of energy being delivered by the laser per second, we
can calculate the number of photons being emitted from their quotient.
Ephoton = hν
λν = c
hc
Ephoton =
λ
6.626 × 10−34 Js 3.00 × 108
Ephoton =
405 × 10−9 m
Ephoton = 4.90 × 10−19 J
m
s
In 1 second, if we emit at a rate of 20 mW, the energy emitted is 20 mW × 1 s = 20 mJ .
This must be caried by n photons, such that n × Ephoton = Etotal . Thus:
Etotal
Ephoton
n = 20 × 10−3 J /4.90 × 10−19 J
n=
n = 4.08 × 1016 photons each second
Because only 1.00 % of the photons contribute to the photoelectric effect, and each of those
photons which does contribute ejects a single electron, we get 4.08 × 1014 e− per second.
This is not a lot, but is very easily measurable with an regular ammeter.
A 1999 article1 in the journal Nature describes the experimental detection of the diffraction
pattern produced by de Broglie waves of C60 (buckyballs).
The diameter of a C60 molecule is approximately 1.0 nm. Assuming the C60 molecule is
traveling at 220 m/s , what is the ratio of its diameter to de Broglie wavelength?
h
p
p = mv
h
λ=
mv
λ=
60 × 12.01 g
1 kg
1 mol
×
×
23
6.022 × 10 molecules
1 mol
1000 g
−24
m = 1.20 × 10
kg
6.626 × 10−34 Js
λ=
1.20 × 10−24 kg × 220 m/s
λ = 2.51 × 10−12 m = 2.51 pm
m = 1 molecule ×
d
= 1.0 nm /2.51 pm
λ
= 400
1 Arndt,
M. et al. Wave-particle duality of C60 molecules. Nature 401, 680682 (1999)
The equations and potential energy functions used in quantum mechanics generally disregard
gravity and focus on only electromagnetic forces.
Recall that the gravitational and electrostatic forces are governed by similar inverse square
1 q1 q2
laws: FG = G m1r2m1 , FC = 4π
2 .
0 r
Calculate the ratio of the electrostatic and gravitational forces between the electron and
nucleus in hydrogen.
Based on your answer, do you think the disregard of gravity in quantum mechanics is valid?
1
q q
1 2
FC
r2
= 4πm0 1 m
FG
G r2 1
1
q1 q2
=
4π0 G m1 m2
1
1
=
2
−1
−12
4π0 G
4π · 8.854 × 10
C N m−2 · 6.67 × 10−11 N m2 kg−2
= 1.34 × 1020 kg2 / C2
FC
(1.602 × 10−19 C )2
20
2
2
= 1.34 × 10 kg / C
FG
(9.11 × 10−31 kg )(1.67 × 10−27 kg )
= 2.26 × 1039
Because the electrostatic force utterly swamps the gravitational force, the disregard of gravity
in the quantum mechanical equations is valid.
β-Carotene is an orange colored compound that primarily responsible for the orange color
of carrots and is a precursor for the biosynthesis of vitamin A.
Figure 1: Molecular Structure of β-Carotene
The long strand of alternating double and single bonds allows the π electrons in β-Carotene
electrons delocalize over the entire length of the chain. Because the molecule is psuedo 1dimensional, it can be reasonably described using the particle-in-a-box model.
The maximum absorption of light by β-Carotene occurs at a wavelength of 480 nm. If this
absorption corresponds to a an n = 11 to n = 12 transition of an electron in a particle in a
box system, calculate the length of the box.
Is this a physically reasonable value for the length of the β-Carotene molecule?
c = λν
∆E = hν
hc
∆E =
λ
2 2
hn
En =
8mL2
∆E11→12 =
s
L=
h2
(122 − 112 )
8mL2
hc
23h2
=
λ r8mL2
23hλ
L=
8mc
23 · 6.626 × 10−34 Js
8 · 9.11 × 10−31 kg · 3.00 × 108 m/s
L = 1.8 × 10−9 m = 1.8 nm = 18 Å
A carbon-carbon bond is about 1.5 Å, so the ∼20 C-C bonds that make up the chain of
β-Carotene are on the order of 30 Å. I’d say this is very close to our calculated length of 18
Ågiven the simplicity of the model.
Conceptual Questions
1.
“Those who are not shocked when they first come across quantum theory cannot possibly have understood it.” - Niels Bohr
What do you think are the most shocking things about QM, or the parts that are most
different from classical mechanics?
2. Why do uncertainty in position and momentum trade off against each one another?
Does the uncertainty principle apply to other waves like sound and water waves?
3. A general rule of thumb is that for a wave function, the number of nodes is directly
related to the energy (more nodes → higher energy). Why is this true?
4. Briefly describe one a phenomena or technology that you experience or use in your daily
life that is inherently quantum mechanical and could not be explained or understood
in classical terms.