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Past exam of the mathematics course of the University of Cambridge
/
2025
/
ii
/
Paper 3
/
14E
/
b
/
iii
/
Solution
...
2025
ii
Paper 3
14E
b
iii
OurBigBook.com
Words: 66
Define
K
(
T
)
=
n
B
1
(
h
2
2
π
m
e
k
B
T
)
3/2
=
8
η
ζ
(
3
)
2
π
(
k
B
T
m
e
c
2
)
3/2
.
(106)
Using
n
e
=
n
B
F
,
n
j
+
=
X
j
+
n
j
, and
n
j
0
=
(
1
−
X
j
+
)
n
j
, the two Saha relations become the closed system
F
1
−
X
H
+
X
H
+
=
K
(
T
)
e
−
I
H
/
(
k
B
T
)
,
(107)
F
1
−
X
H
e
+
X
H
e
+
=
2
K
(
T
)
e
−
I
He
/
(
k
B
T
)
,
(108)
where
F
=
(
1
−
Y
p
)
X
H
+
+
4
Y
p
X
H
e
+
.
(109)
These are two equations for the two ionization fractions.
When
Y
p
→
0
,
F
=
X
H
+
, and the hydrogen equation reduces to
1
−
X
H
+
X
H
+
2
=
n
B
1
(
h
2
2
π
m
e
k
B
T
)
3/2
e
−
I
H
/
(
k
B
T
)
,
(110)
the usual
hydrogen-only Saha equation
.
Solved by gpt-5.6-sol high.
Ancestors
(12)
Iii
B
14E
Paper 3
Ii
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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