South Pasadena • AP Chemistry
Name
8 ▪ Atomic Theory
Period
8.1
PROBLEMS
1. De Broglie Wavelength. Calculate the de Broglie
wavelength for each of the following:
(1 J = 1 kg·m2/s2)
a. An electron with a velocity 10% of the speed
of light. (me− = 9.11 × 10−31 kg)
h
λ=
m·v
(6.626 × 10−34 J·s)
=
(9.11 × 10−31 kg)(0.1)(3.0 × 108 m/s)
= 2.42 × 10−11 m
b. A tennis ball (0.055 kg) served at 35 m/s.
h
(6.626 × 10−34 J·s)
λ=
=
m·v (0.055 kg) (35 m/s)
= 3.44 × 10−34 m
c. Explain why the wavelength of objects are not
observed macroscopically.
The wavelength is too short to be observed.
2. Photoelectric Effect. When purple light with a
wavelength of 349 nm is directed at an iron
surface, electrons are ejected and found to travel at
7.85 × 105 m/s. (1 nm = 10−9 m)
a. What is the kinetic energy of these ejected
electrons? (me− = 9.11 × 10−31 kg,
1
KE = 2 m·v2)
1
KE = (9.11 × 10−31 kg)(7.85 × 105 m/s)2
2
= 2.81 × 10−19 J
b. What is the energy of the light used?
h·c (6.626 × 10−34 J·s)(3.0 × 108 m/s)
E=
=
λ
(349 × 10−9 m)
= 5.70 × 10−19 J
c. What is the energy with which these electrons
are bound to the atom? (Ebinding = Elight – KE)
Ebinding = (5.70 × 10−19 J) – (2.81 × 10−19 J)
= 2.89 × 10−19 J
–
QUANTUM
Date
MECHANICS
d. Would light with a wavelength of 400 nm be
able to eject these electrons? Explain with
calculations.
h·c (6.626 × 10−34 J·s)(3.0 × 108 m/s)
E=
=
λ
(400 × 10−9 m)
= 4.97 × 10−19 J
Yes, because this light has more energy
than the Ebinding.
3. Hydrogen Emission Spectrum. The energy for
the nth shell of the hydrogen atom is given by En =
−2.178 × 10−18
J. Calculate the energy and
n2
color/type of light emitted between the following
transitions.
−2.178 × 10−18
E4 =
J = −1.40 × 10−19 J
42
−2.178 × 10−18
E3 =
J = −2.40 × 10−19 J
32
−2.178 × 10−18
E2 =
J = −5.40 × 10−19 J
22
a. n = 4 to n = 3
E3 – E4 = (−2.40 × 10−19 J) – (−1.40 × 10−19 J)
= −1.00 × 10−19 J
b. n = 4 to n = 2
E2 – E4 = (−5.40 × 10−19 J) – (−1.40 × 10−19 J)
= −4.00 × 10−19 J
c. n = 3 to n = 2
E2 – E3 = (−5.40 × 10−19 J) – (−2.40 × 10−19 J)
= −3.00 × 10−19 J
d. At higher levels, does the energy difference
between shells increase or decrease or remain
the same?
The difference of energy between shells
decreases at higher levels.
4. Quantum Numbers. Identify the n and ℓ quantum
numbers for the following subshells. Then state
the possible values for the mℓ quantum number.
a. 6p
n = 6, ℓ = 1, mℓ = −1, 0, +1
b. 3s
n = 3, ℓ = 0, mℓ = 0
c. 4d
n = 4, ℓ = 2, mℓ = −2, −1, 0, +1, +2
d. 5p
n = 5, ℓ = 1, mℓ = −1, 0, +1
5. Quantum Numbers. Determine whether the
following sets of quantum numbers are allowed.
If so, state what subshell it describes.
a. (4, 2, −1) 4d
b. (5, 0, −1) Not allowed
c. (4, 4, −1) Not allowed
d. (5, 4, +5) Not allowed
7. Atomic Orbitals. Describe the difference in
shape between the following orbitals.
a. 2s vs. 2p
2s orbital is spherical and has a spherical
node; 2p orbital is dumbbell-shaped and
has a planar node.
b. 1s vs. 2s
1s and 2s orbitals are both spherical. 1s
does not have a node, and 2s orbital has a
spherical node.
c. 2px vs. 2py
2px and 2py are both dumbbell-shaped, but
are oriented in different directions.
6. Quantum Numbers. How many electrons can
have the following quantum numbers in an atom.
a. n = 2, ℓ = 1
6 electrons
b. n = 4, ℓ = 2, mℓ = −2
2 electrons
c. n = 7
98 electrons
d. n = 5, ℓ = 3, mℓ = −1
2 electrons
AP Chemistry 2007B #2
Answer the following problems about gases.
(a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two
Isotope Mass (amu)
common isotopes of naturally occurring neon as indicated in the table below.
Ne-20
19.99
(i) Using the information above, calculate the percent abundance of each isotope.
(19.99)(x) + (21.99)(1 – x) = 20.18
Ne-22
21.99
x = 0.905
Ne-20: 90.5%, Ne-22: 9.5%
(ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon.
1 mol Ne  6.02 × 1023 atoms Ne 9.5 atoms Ne-22
22
12.55 g Ne
1 mol Ne
20.18 g Ne 
  100 atoms Ne  = 3.56 × 10 atoms Ne-22
(b) A major line in the emission spectrum of neon corresponds to a frequency of 4.34 × 1014 s–1. Calculate the
wavelength, in nanometers, of light that corresponds to this line.
c 3.0 × 108 m/s  1 nm 
λ= =
= 691 nm
ν 4.34 × 1014 s−1 10−9 m
(c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as shown by
the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the Sun.
uv
O3 (g) →
O2 (g) + O (g)
A molecule of O3 (g) absorbs a photon with a frequency of 1.00 × 1015 s–1.
(i) How much energy, in joules, does the O3 (g) molecule absorb per photon?
E = h·ν = (6.626 × 10−34 J·s)(1.00 × 1015 s−1) = 6.626 × 10−19 J
(ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol–1. Does a photon
with a frequency of 1.00 × 1015 s–1 have enough energy to break this bond? Support your answer with a
calculation.
387 kJ1000 J
1 mol
 = 6.42 × 10−19 J
1 mol  1 kJ 6.02 × 1023 bonds
Since 6.626 × 10−19 J > 6.42 × 10−19 J, the photon has enough energy to break the bond.