Version 232 – Exam 1 – McCord – (53130)
1
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printout
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have
30
questions.
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before answering.
McCord
CH301tt
This exam is only for
McCord’s Tues/Thur CH301 class.
c
= 3
.
00
×
10
8
m/s
h
= 6
.
626
×
10

34
J
·
s
m
e
= 9
.
11
×
10

31
kg
N
A
= 6
.
022
×
10
23
mol

1
ν
=
R
1
n
2
y

1
n
2
x
where
R
= 3
.
29
×
10
15
s

1
ψ
n
(
x
) =
2
L
1
2
sin
n
π
x
L
E
n
=
n
2
h
2
8
mL
2
n
= 1
,
2
,
3
,
· · ·
001
10.0 points
A quantum mechanical particle in a box in its
ground state is most likely to be found
1.
in the middle of the box.
correct
2.
is equally likely to be found at all positions
except the very center of the box.
3.
at both edges of the box.
4.
is equally likely to be found at all positions
in the box.
5.
at the very edge of the box.
Explanation:
The ground state is
Ψ
1
which has maximum
probability in the middle of the box.
002
10.0 points
The graph shows the radial distribution plots
for the 1s wavefunctions for H, He, and He
+
.
A
B
C
radius
4
π
r
2
Ψ
2
Which plot is the 1s wavefunction for the He
+
ion?
1.
There is no way to know
2.
C
3.
A
correct
4.
B
Explanation:
H has one electron and one proton. He has
two electrons and two protons. He
+
has one
electron and two protons. Therefore the elec
tron will have the greatest attraction to the
the nuclei with two protons. In He there will
be a slight reduction in the e
ff
ective nuclear
charge due to the electronelectron repulsion.
In He
+
, there is only one electron so the radius
will be the smallest. Plot A is peaked closest
to the nucleus so it will be the smallest.
003
10.0 points
What are the correct quantum numbers (
n
,
,
m
) for the seventh electron of Cl?
1.
2, 1, 1
correct
2.
2, 2, 1
3.
3, 2, 1
4.
3, 1, 0
5.
2, 1, 2
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Version 232 – Exam 1 – McCord – (53130)
2
6.
3, 1, 1
7.
2, 0, 1
8.
3, 0, 0
Explanation:
The electron configuration for Cl is
1
s
2
2
s
2
2
p
6
3
s
2
3
p
5
The seventh electron is in a 2
p
orbital, so
n
= 2,
= 1,
m
=

1, 0, or +1.
2, 1, 0 and 2, 1,

1 are not given as choices.
004
10.0 points
Carbon emits photons at 745 nm when ex
posed to blackbody radiation.
How much
energy would be obtained if 44 g of carbon
were irradiated?
1.
2
.
67
×
10

19
J
2.
7
.
08
×
10
6
J
3.
9
.
11
×
10

21
J
4.
1
.
17
×
10

17
J
5.
7
.
08
×
10
3
J
6.
5
.
90
×
10
5
J
correct
Explanation:
λ
= 745 nm = 7
.
45
×
10

7
m
m
C
= 44 g
Assume each carbon atom emits one pho
ton. For each photon
E
1
=
h
ν
=
h c
λ
=
(6
.
626
×
10

34
J
·
s) (3
×
10
8
m
/
s)
7
.
45
×
10

7
m
= 2
.
66819
×
10

19
J
,
so the total energy emitted is
E
=
E
1
n
=
2
.
66819
×
10

19
J
C atom
×
6
.
022
×
10
23
C atoms
1 mol C atoms
×
1 mol C atoms
12
.
01 g C
×
(44 g C)
= 5
.
88663
×
10
5
J
.
005
10.0 points
How
many
electrons
can
possess
this
set
of
quantum
numbers:
principal
quantum
number
n
= 4, magnetic quantum number
m
=

1?
1.
14
2.
12
3.
16
4.
8
5.
0
6.
4
7.
2
8.
18
9.
10
10.
6
correct
Explanation:
Use the rules for the quantum numbers:
If
n
= 4 then
= 0
,
1
,
2
,
3; however, for
m
=

1,
= 1
,
2
,
3. Each of these permit
ted sets of values of
n
,
and
m
specifies ONE
orbital:
n
= 4,
= 1,
m
=

1:
4
p
n
= 4,
= 2,
m
=

1:
4
d
n
= 4,
= 3,
m
=

1:
4
f
and each orbital can have
m
s
=
±
1
2
;
i.e.
, can
hold two electrons.
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 Spring '08
 wandelt
 Chemistry, Atom, pH, Atomic orbital, He+

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