dewitt (aad777) – Homework 2 – sutcliffe – (52380)
1
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printout
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have
20
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
Starting in the ground state n = 1 in a hydro
gen atom, an electron is excited to a higher
energy level by absorption of 121.6 nm light.
What is the final energy state of the electron?
1.
n = 4
2.
n = 3
3.
n = 7
4.
n = 5
5.
n = 2
correct
6.
n = 6
Explanation:
121.6 nm is a frequency of 2
.
47
×
10
15
s

1
.
That corresponds to a transition from n =1
to n =2 using the Rydberg formula.
002
10.0 points
If a particle is confined to a onedimensional
box of length 300 pm, for Ψ
3
the particle is
most likely to be found at
1.
17.3 pm.
2.
100 and 200 pm, respectively.
3.
50, 150, and 250 pm, respectively.
cor
rect
4.
0 pm.
5.
300 pm.
Explanation:
003
10.0 points
A quantum mechanical particle in a box in its
ground state is most likely to be found
1.
is equally likely to be found at all positions
except the very center of the box.
2.
at the very edge of the box.
3.
in the middle of the box.
correct
4.
at both edges of the box.
5.
is equally likely to be found at all positions
in the box.
Explanation:
The ground state is Ψ
1
which has maximum
probability in the middle of the box.
004
10.0 points
Set(s) of possible values of
m
ℓ
are
A)

4;

3;

2;

1; 0; +1; +2; +3; +4
B)

3;

2;

1; 0; +1; +2; +3
C)

2;

1; 0; +1; +2
D)

1; 0; +1
E) 0
Select the best choice for
n
= 3.
1.
only B
2.
only E
3.
only C
4.
only C, D, and E
correct
5.
only D and E
6.
only B, C, D, and E
7.
A, B, C, D, and E
8.
only A
9.
only D
Explanation:
For any value of
n
, the possible values of
ℓ
range from 0 up to (
n

1).
In turn, for
any value of
ℓ
, the possible values of
m
ℓ
range
from =

ℓ
, ..., 0, ..., +
ℓ
. Since
n
= 3,
ℓ
can
be 0, 1, or 2. Therefore
when
ℓ
= 2, C)
m
ℓ
=

2;

1; 0; +1; +2
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dewitt (aad777) – Homework 2 – sutcliffe – (52380)
2
when
ℓ
= 1, D)
m
ℓ
=

1; 0; +1
when
ℓ
= 0, E)
m
ℓ
= 0
005
10.0 points
The three quantum numbers for an electron in
a hydrogen atom in a certain state are
n
= 4,
ℓ
= 2,
m
ℓ
= 1.
The electron is located in
what type of orbital?
1.
3
d
2.
3
p
3.
4
s
4.
4
d
correct
5.
4
p
Explanation:
The notation is
n
ℓ
, where
n
= 1
,
2
,
3
,
4
,
5
, ...
,
ℓ
= 0
,
1
,
2
, ...,
(
n

1)
represented as a letter:
ℓ
0
1
2
3
4
5
etc.
orbital
s
p
d
f
g
h
and
m
ℓ
=

ℓ,

(
ℓ

1)
,

(
ℓ

2)
, ...,
0
, ...,
+(
ℓ

2)
,
+(
ℓ

1)
, ℓ
.
The value of
m
ℓ
is not needed to determine
the orbital type, as long as it is valid.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, Atom, Electron, Periodic Table, DeWitt

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