The document discusses calculating the vibrational energy of carbon monoxide (CO) using different isotopes of carbon and oxygen. It first calculates the vibrational energy of normal CO to be 0.2691 eV. It then calculates: (a) The vibrational energy would be 0.2596 eV if the oxygen was replaced with the isotope of mass 18.00 u. (b) The vibrational energy would be 0.25542 eV if the carbon was replaced with the radioactive isotope of mass 14.00 u.

1. 1
The vibrational energy of CO is 0.2691 eV
when the constituents of the molecule are
the most abundant isotopes of carbon (m =
12.00 u) and oxygen (m = 16.00 u).
(a) What would be the vibrational energy if
the oxygen were replaced by the less
abundant isotope with m = 18.00 u?
(b) What would be the vibrational energy if
the carbon in the original CO were replaced
with radioactive carbon (used in radiocarbon
dating) with mass 14.00 u?
Helping Tools
No.1
No.2

2. 2
No.3

3. 3
OR

4. 4
Solution

5. 5
1 2
1 2
12 16
12 16
12 16
12 16
2
Let us first calculate the reduced
mass using the relation
of CO made with C and O as
( ) ( )
( )
( ) ( )
(12 )(16 ) 192
6.86
12 16 28
r
m m
m
m m
m C m O
m C O
m C m O
u u u
u
u u u
=
+
=
+
= = =
+

6. 6
1 2
1 2
12 18
12 18
12 18
12 18
2
Let us now calculate the reduced
mass using the relation
of CO made with C and O as
( ) ( )
( )
( ) ( )
(12 )(18 ) 216
7.2
12 18 30
r
m m
m
m m
m C m O
m C O
m C m O
u u u
u
u u u
=
+
=
+
= = =
+

7. 7
15
15
We know that Vibrational
Energy is given as:
(1)
Using
k=1860 & 0.66 10 .
in eq.(1), we get
1860
E 0.66 10 . (2)
Using
7.2 & 7.47 one by one, we get
vib
r
vib
r
r
k
E
m
N
X eV s
m
N
m
X eV s
m
m u u
−
−
= − − − − − − − − − −
=
= − − −
=

8. 8
15
7.2
15
2
2
15
2
16
2
15
16
15
15
1860
0.66 10 .
7.2
1860
0.66 10 .
7.2 931.5
1860
0.66 10 .
7200 931500
1860 9 10
0.66 10 .
7200 931500
1860 9 10
0.66 10
7200 931500
1
0.66 10
u
N
m
E X eV s
u
N
m
X eV s
MeV
X
c
NXc
X eV s
X meV
m
NX X
s
X eV s
X meV
NX X m
X eV
X eV
X eV
−
−
−
−
−
−
=
=
=
=
=
=
10
15
6740000000 10
6706800000
0.66 10 24959740000 /
X Nm
eV
X eV Nm eV
−
=

9. 9
10
15
7.2
15
15
19
15
19
15 19
15 9
16740000000 10
0.66 10
6706800000
0.66 10 24959740000 /
0.66 10 157986.52
1.602 10
1
104271 10
1.602 10
104271 10 0.62 10
104271 10 2.49 10
259634.7
u
X Nm
E X eV
eV
X eV Nm eV
J
X eVX
X J
eVX X
X
eVX X X
eVX X X
−
−
−
−
−
−
−
−
 =
=
=
=
=
=
= 6
9 10 0.2596
X eV eV
−
=

10. 10
1 2
1 2
14 16
14 16
14 16
14 16
2
(b) Now, let us calculate the reduced
mass using the relation
of CO made with C and O as
( ) ( )
( )
( ) ( )
(14 )(16 ) 224
7.47
14 16 30
r
m m
m
m m
m C m O
m C O
m C m O
u u u
u
u u u
=
+
=
+
= = =
+

11. 11
15
7.47
15
2
2
15
2
16
2
15
16
15
15
1860
0.66 10 .
7.47
1860
0.66 10 .
7.47 931.5
1860
0.66 10 .
7470 931500
1860 9 10
0.66 10 .
7470 931500
1860 9 10
0.66 10
6958305000
0.66 10
u
N
m
E X eV s
u
N
m
X eV s
MeV
X
c
NXc
X eV s
X meV
m
NX X
s
X eV s
X meV
NX X m
X eV
eV
X e
−
−
−
−
−
−
=
=
=
=
=
=
10
15
16740000000 10
6958305000
0.66 10 24057582988 /
X Nm
V
eV
X eV Nm eV
−
=

12. 12
15 19 1
7.47
15 28 15 14
1
1
0.66 10 24057582988 10
1.602
0.66 10 15.02 10 0.66 10 3.87 10
2.5542 10 0.25542
u
E X eV NmX X J
X eV X X eVX X
X eV eV
− −
− −
−
 =
= =
= =