This preview shows pages 1–5. Sign up to view the full content.
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 = 12
Wrong units = 1
2 points
1/2 for significant figures
3 pts
2 points, 1 each
1/2 each for significant figures
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentAnswer 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.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentAnswer Key
Ch 1a, Problem Set Two
Side 4 of 9
Problem Three – Range of Colors – 5 Points
Final Answer: For Helium ion, the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.
 Fall '08
 Lewis

Click to edit the document details