what is the population ratio of two molecular energy states i and j,if k=1/2,ei=kbtat 27°c
Answers
Answer:
hi how are you
Answer:
This can be gotten from statistical mechanics. The fraction of occupied states is:
N
i
N
=
g
i
e
−
β
ε
i
∑
k
g
k
e
−
β
ε
k
where:
N
i
is the number of molecules populating state
i
, and
N
is the total number of molecules.
q
i
=
g
i
e
−
β
ε
i
is the partition function for state
i
.
g
i
is the degeneracy of state
i
.
β
=
1
k
B
T
is a constant.
k
B
=
1.38065
×
10
−
23
J/K
is the Boltzmann constant.
T
is the temperature in
K
.
ε
i
is the energy of state
i
.
At a single temperature
T
, the ratio of the populations of states
i
and
j
is:
N
i
/
N
N
j
/
N
=
N
i
N
j
=
g
i
e
−
β
ε
i
∑
k
g
k
e
−
β
ε
k
g
j
e
−
β
ε
j
∑
k
g
k
e
−
β
ε
k
=
g
i
e
−
β
ε
i
g
j
e
−
β
ε
j
This ratio,
N
i
N
j
, was given as
1.059
×
10
−
30
. This physically means that very little of the metastable state is actually occupied.
When the upper state
i
corresponds to a metastable state
i
, it indicates that
ε
i
>
ε
j
, since a metastable state is basically a temporarily stable state, and is thus more unstable than the ground state. Dividing through gives:
=
g
i
g
j
e
−
β
ε
i
−
(
−
β
ε
j
)
Factor out the negative beta in the exponent to get:
N
i
N
j
=
g
i
g
j
e
−
β
[
ε
i
−
ε
j
]
=
g
i
g
j
e
−
β
Δ
ε
Assuming we are talking about singly-degenerate electronic states, we therefore assume that the degeneracies are both
1
, so that
g
i
=
g
j
=
1
.
Then, the energy is analogous to the one you had been taught from general chemistry:
Δ
ε
=
h
ν
=
h
c
λ
Therefore:
N
i
N
j
=
e
−
Δ
ε
/
k
B
T
=
e
−
h
c
/
λ
k
B
T
Now, we can solve for the wavelength.
ln
(
N
i
N
j
)
=
−
h
c
λ
k
B
T
λ
ln
(
1.059
×
10
−
30
)
=
−
69.02
λ
=
−
h
c
k
B
T
⇒
λ
=
h
c
69.02
k
B
T
=
(
6.626
×
10
−
34
J
⋅
s
)
(
2.998
×
10
8
m/s
)
69.02
(
1.38065
×
10
−
23
J/K
)
(
330 K
)
=
6.317
×
10
−
7
m
This would be then, about
631.7 nm
.