Answer Key
Ch 1a, Problem Set Two
Side 1 of 9
Problem One – EM Radiation – 15 Points, 5 Each
a.
(5 Points)
E
hc
hc
h
E
=
!
=
=
"
"
#
"
=
6.626
#
10
$
34
J
%
s
(
)
2.998
#
10
8
m
s
(
)
10
9
nm
1
m
&
’
(
)
*
+
3.61
#
10
$
19
J
=
550.27nm
Final Answer: 550 nm, Yellow
b.
(5 Points)
!
!
"
#
$
$
%
&
’
(
=
’
=
)
2
1
2
1
1
1
*
*
*
*
hc
hc
hc
E
"
E
=
6.626
#
10
$
34
J
%
s
(
)
2.998
#
10
8
m
s
(
)
1
365.0
nm
$
1
600.5
nm
&
’
(
)
*
+
10
9
nm
m
&
’
(
)
*
+
1
eV
1.602
#
10
$
19
J
&
’
(
)
*
+
=
1.332
eV
Final Answer: 3.546 eV
c.
(5 Points)
"#
=
c
$
"
=
c
#
=
2.998
%
10
8
m
s
89.3
%
10
6
s
&
1
=
3.3572 meters
E = h
ν
=
6.626
"
10
#
34
J
$
s
"
89.3
"
10
6
s
#
1
=
7.06994x10
#
26
J
Final Answer: KROQ’s wavelength is 3.36 m with E = 5.92 x 10
-26
J
2 pts
2 pts for correct wavelength
1 point for correct color (orange ok)
3 pts
Sig fig = -1/2
Calculation error = 1-2
Wrong units = -1
2 points
-1/2 for significant figures
3 pts
2 points, 1 each
-1/2 each for significant figures
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Answer Key
Ch 1a, Problem Set Two
Side 2 of 9
Problem Two – Bohr Model – 15 Points
Write Rydberg Equation in terms of wavelength difference we’ll call
"
fi
:
For Hydrogen atom
E
=
h
"
=
hc
#
fi
=
hcR
H
Z
2
1
n
f
2
$
1
n
i
2
%
&
’
’
(
)
*
*
"
fi
=
1
R
H
1
n
f
2
#
1
n
i
2
$
%
&
’
(
)
#
1
For Helium
E
=
h
"
=
hc
#
fi
=
hcR
H
Z
2
1
n
f
2
$
1
n
i
2
%
&
’
’
(
)
*
*
"
fi
=
1
4R
H
1
n
f
2
#
1
n
i
2
$
%
&
&
’
(
)
)
#
1
The leading coefficient of this equation is a constant that is easily calculated:
1
4R
H
=
22.8
nm
Now we must move systematically through each
n
i
value of each
n
f
until we have
exhausted the possibility of finding visible light…for simplicity, use the series discussed
in class…
n
f
n
i
Lyman Lines:
n
f
=1,
n
i
≥
2
Location
1
2
91.2
nm
(
)
"
1
1
2
#
1
2
2
$
%
&
’
(
)
#
1
=
30.5
nm
UV
1
3
91.2
nm
(
)
"
1
1
2
#
1
3
2
$
%
&
’
(
)
#
1
=
25.75
nm
UV
1
4
91.2
nm
(
)
"
1
1
2
#
1
4
2
$
%
&
’
(
)
#
1
=
24.33
nm
UV
1
5
91.2
nm
(
)
"
1
1
2
#
1
5
2
$
%
&
’
(
)
#
1
=
23.75
nm
UV
n
f
n
i
Balmer Lines:
n
f
=2,
n
i
≥
3
Location
2
3
91.2
nm
(
)
"
1
2
2
#
1
3
2
$
%
&
’
(
)
#
1
=
164
nm
UV
2
4
91.2
nm
(
)
"
1
2
2
#
1
4
2
$
%
&
’
(
)
#
1
=
122
nm
UV
2
5
91.2
nm
(
)
"
1
2
2
#
1
5
2
$
%
&
’
(
)
#
1
=
109
nm
UV
n
f
n
i
Paschen Lines:
n
f
=3,
n
i
≥
4
Location
3 pts for correct
derivation
3 pts for correct
derivation
Answer Key
Ch 1a, Problem Set Two
Side 3 of 9
3
4
91.2
nm
(
)
"
1
3
2
#
1
4
2
$
%
&
’
(
)
#
1
=
469
nm
Visible -Blue
3
5
91.2
nm
(
)
"
1
3
2
#
1
5
2
$
%
&
’
(
)
#
1
=
321
nm
UV
All other transitions will therefore be in the visible or higher wavelengths.
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