Problem Set 2_Sol - Answer Key Side 1 of 9 Ch 1a Problem...

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

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.