Quantum Mechanics of Hydrogen Molecule Rotations

 
Rotational heat capacity of
Hydrogen molecule
 
 
 
Moment of inertia of hydrogen very small so
rotational quanta too large
so classical treatment not used
rotational partition function calculated only by
summation
But calculated value does not agree with
experimental value
 
 
 
Difference is due to two nuclear spin isomers
for hydrogen molecule
Nuclear spin of the two protons can be paired
to get either symmetric combination (spin
parallel) or antisymmetric combination (spin
opposed)
Former is ortho and latter para hydrogen
 
 
Wave number 
ν
= hJ/4
π
2
Ic for rotational space of
diatomic molecule
Spectra of ortho hydrogen has fairly intense lines
corresponding to odd values of J
Para hydrogen has less intense lines
corresponding to even value of J
Spectra of hydrogen has alternating intense and
less intense rotational lines – shows that mixture
contains ortho(75%) and para hydrogen
 
 
If system has non-interacting particles, wave
function for over all system is product of wave
function of all particles
For particles 1 and 2
a
(1) or a(1) , 
b
(2) or b(2)
For combination 
I
 = 
 
a
(1)
b
(2) = a(1)b(2)
 
 
If indistingushiable particles, same wave
function can be available for both
So 
II
 = 
a
(2) 
b
(1)  = a(2)b(1) is also valid
But by quantum mechanics this is not
acceptable
 
 
To correct this – use in phase(+) and out
phase(-) combinations of 
1
 and 
2
 for in phase = symmetric= 
s
Out phase = antisymmetric = 
a
s
 = c’(
I
 +
II
) = c’[a(1)b(2) + a(2)b(1)
a
 = c’(
I
 - 
II
) = c’[a(1)b(2) – a(2)b(1)
C’ is normalising constant = 1/
 2
 
 
Interchanging of labels leaves 
s
 unchanged
and 
a
 becomes - 
a
E
total
 = E
e
 + E
t
 + E
v
 + E
r
 + E
n
total
 = 
e
 + 
t
 + 
v
 + 
r
 + 
n
If sum of numbers of proton , electron and
neutron = P , is even
 must be symmetric and only 
s
 is allowed
If P is odd, antisymmetric- 
a
 is allowed
 
 
Symmetry of 
 on exchange will depend on
whether exchange of nuclei 0r electrons or
both
If only nuclei exchanged – effects of electron
exchange not considered – but effect of
nuclear exchange on 
e
 can be considered
r
 is characterised by quantum number J ,
each J value is associated with (2J + 1)
function
 
 
r
 is symmetric if J is even
      antisymmetric if J is odd
r
 is a function of angles 
 and 
π
Nuclear exchange means changing 
 to (
-
π
)
and 
 to (
+
π
)
Hydrogen has odd number of particles so
antisymmetric
 
 
n depends on inter nuclear separation so
nuclear exchange has no effect
n is therefore symmetric
It hydrogen in electronic ground state 
e has
symmetric 
g
+
 ,
+ means function is symmetric and symmetric
w.r.t inversion as given by g
So over all wave function is symmetric
 
 
If H-H is assumed as A-B, each has spin+1/2 or -
1/2
The possible combination is
A(+)B(+) + B(+)A(+)  = A(+)B(+)
A(-)B(-)  +  B(-)A(-)    = A(-)B(-)
A(+)B(-)  + B(+)A(-)
A(-)B(+)   + B(-)A(+) both 3 &4 are similar
These are symmetric nuclear spin function and
has three different function.
 
 
A(+)B(-) - B(+)A(+)  = 0
A(-)B(-)  -  B(-)A(-)    = 0
A(+)B(-)  - B(+)A(-)
A(-)B(+)   - A(-)B(+) both 3 &4 are similar –
these are antisymmetric and has only one
function
1 & 2 are excluded
 
 
So number of symmetric wave function is 3
times more than antisymmetric wave function
In general – homo nuclear diatomic – of spin
‘i’ – ratio of symmetric to antisymmetric spin
function is (i+1) : I
In general – one with greater value is ortho
and the other is para
 
 
It is possible to find combinations of wave
function which make 
total 
either symmetric or
antisymmetric
e
, 
v
 and 
t
 are symmetric – so overall
symmetry determined by products of 
r 
x 
n
For hydrogen total wave function must be
antisymmetric since ‘P’ is odd
total
 to be antisymmetric either 
r
 or 
n
 must
be antisymmetric but not both
 
 
r
 symmetric J is even
n
 must be antisymmetric(s=0)
r
 antisymmetric J odd
n
 symmetric (s=1)
For pare J is even, s=0
Ortho J is odd , s= 1
 
 
In general – molecule in ortho state occupy
odd rotational level- with pare occupy even
rotational level
Єr =BJ(J+1)    B= h
2
/8
π
2
I
So f
r
(ortho) =
             (2s+1)
s=1
 
J=1,3,5..
(2J+1)e-
BJ(J+1)/kT
     …..(1)
Since each state has nuclear statistical weight
(2s+1), s=0 for para and s=1 for ortho
 
 
So f
r
(para) =
             (2s+1)
s=0
 
J=0,2,4..
(2J+1)e-
BJ(J+1)/kT
     …..(2)
From (2)
 as T→0 , all molecules are populated with J=0 ,
so no ortho form- not easy to convert ortho to
para – without catalyst or other reagent
 
 
So f
r
(ortho) =
                      3
J=1,3,5..
(2J+1)e-
BJ(J+1)/kT
So f
r
(para) =
                      1 
J=0,2,4..
(2J+1)e-
BJ(J+1)/kT
     …..(1)
Relative numbers of ortho to para forms of
hydrogen will be in their ratio of their
partition function
At 300K, -B/kT = 0.3
 
 
Summation of ortho series is
3e
-0.6
 +7e
-3.6
 +11e
-9
 +….
For para
1+5e
-1.8
 +9e
-6
 +….
For ortho series sum = 1.7
For para sum = 1.8
 
 
At higher temperatures two become similar
n
o
/n
p
 = f
o
/f
p
 =(2s+1)
0
/(2s+1)
p
  = 3/1
ie 25% para and 75% ortho
To calculate Cv, evaluate f for ortho and para
Rotational heat capacity of ordinary H
2
 is thus
3/4Cv
(ortho)
 + 1/4Cv
(para)
 at a given temperature
 
 
At low temperatures , H
2
 molecules are in
lowest possible rotational level
Є
r
 = J(J+1)h
2
/8
π
2
I
Since J=1 for ortho
f
r(ortho)
 = 9e
-h2/4
π
2I
For para J=o
f
r(para)
 =1e
-0h2/8
π
2I
 =1
Cv = d/dT[RT
2
dlnZ/dT]
v
 
 
= d/dT[LkT
2
dlnf/dT]
Since lnf = constant
Cv(ortho) = 0
Cv(para) = 0
Rotational contribution to heat capacity is zero at
low temperature
As temperature is raised rotational contribution
becomes significance and specific heat increases.
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Dive into the intriguing world of molecular rotations in hydrogen molecules, exploring topics such as rotational heat capacity, moment of inertia, nuclear spin isomers, and wave number calculations. Understand the implications of particle interactions, distinguishable vs. indistinguishable particles, and the importance of symmetry in wave functions. Unravel the complexities of quantum mechanics that govern the behaviors of these fundamental particles.


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  1. Rotational heat capacity of Hydrogen molecule

  2. Moment of inertia of hydrogen very small so rotational quanta too large so classical treatment not used rotational partition function calculated only by summation But calculated value does not agree with experimental value

  3. Difference is due to two nuclear spin isomers for hydrogen molecule Nuclear spin of the two protons can be paired to get either symmetric combination (spin parallel) or antisymmetric combination (spin opposed) Former is ortho and latter para hydrogen

  4. Wave number = hJ/42Ic for rotational space of diatomic molecule Spectra of ortho hydrogen has fairly intense lines corresponding to odd values of J Para hydrogen has less intense lines corresponding to even value of J Spectra of hydrogen has alternating intense and less intense rotational lines shows that mixture contains ortho(75%) and para hydrogen

  5. If system has non-interacting particles, wave function for over all system is product of wave function of all particles For particles 1 and 2 a(1) or a(1) , b(2) or b(2) For combination I= a(1) b(2) = a(1)b(2)

  6. If indistingushiable particles, same wave function can be available for both So II= a(2) b(1) = a(2)b(1) is also valid But by quantum mechanics this is not acceptable

  7. To correct this use in phase(+) and out phase(-) combinations of 1and 2 for in phase = symmetric= s Out phase = antisymmetric = a s= c ( I+ II) = c [a(1)b(2) + a(2)b(1) a= c ( I- II) = c [a(1)b(2) a(2)b(1) C is normalising constant = 1/ 2

  8. Interchanging of labels leaves sunchanged and abecomes - a Etotal= Ee+ Et+ Ev+ Er+ En total= e+ t+ v+ r+ n If sum of numbers of proton , electron and neutron = P , is even must be symmetric and only sis allowed If P is odd, antisymmetric- ais allowed

  9. Symmetry of on exchange will depend on whether exchange of nuclei 0r electrons or both If only nuclei exchanged effects of electron exchange not considered but effect of nuclear exchange on ecan be considered ris characterised by quantum number J , each J value is associated with (2J + 1) function

  10. ris symmetric if J is even antisymmetric if J is odd ris a function of angles and Nuclear exchange means changing to ( - ) and to ( + ) Hydrogen has odd number of particles so antisymmetric

  11. n depends on inter nuclear separation so nuclear exchange has no effect n is therefore symmetric It hydrogen in electronic ground state e has symmetric g+, + means function is symmetric and symmetric w.r.t inversion as given by g So over all wave function is symmetric

  12. If H-H is assumed as A-B, each has spin+1/2 or - 1/2 The possible combination is A(+)B(+) + B(+)A(+) = A(+)B(+) A(-)B(-) + B(-)A(-) = A(-)B(-) A(+)B(-) + B(+)A(-) A(-)B(+) + B(-)A(+) both 3 &4 are similar These are symmetric nuclear spin function and has three different function.

  13. A(+)B(-) - B(+)A(+) = 0 A(-)B(-) - B(-)A(-) = 0 A(+)B(-) - B(+)A(-) A(-)B(+) - A(-)B(+) both 3 &4 are similar these are antisymmetric and has only one function 1 & 2 are excluded

  14. So number of symmetric wave function is 3 times more than antisymmetric wave function In general homo nuclear diatomic of spin i ratio of symmetric to antisymmetric spin function is (i+1) : I In general one with greater value is ortho and the other is para

  15. It is possible to find combinations of wave function which make total either symmetric or antisymmetric e, vand tare symmetric so overall symmetry determined by products of r x n For hydrogen total wave function must be antisymmetric since P is odd totalto be antisymmetric either ror nmust be antisymmetric but not both

  16. rsymmetric J is even nmust be antisymmetric(s=0) rantisymmetric J odd nsymmetric (s=1) For pare J is even, s=0 Ortho J is odd , s= 1

  17. In general molecule in ortho state occupy odd rotational level- with pare occupy even rotational level r =BJ(J+1) B= h2/8 2I So fr(ortho) = (2s+1)s=1 J=1,3,5..(2J+1)e-BJ(J+1)/kT Since each state has nuclear statistical weight (2s+1), s=0 for para and s=1 for ortho ..(1)

  18. So fr(para) = (2s+1)s=0 J=0,2,4..(2J+1)e-BJ(J+1)/kT From (2) as T 0 , all molecules are populated with J=0 , so no ortho form- not easy to convert ortho to para without catalyst or other reagent ..(2)

  19. So fr(ortho) = 3 J=1,3,5..(2J+1)e-BJ(J+1)/kT So fr(para) = 1 J=0,2,4..(2J+1)e-BJ(J+1)/kT ..(1) Relative numbers of ortho to para forms of hydrogen will be in their ratio of their partition function At 300K, -B/kT = 0.3

  20. Summation of ortho series is 3e-0.6+7e-3.6+11e-9+ . For para 1+5e-1.8+9e-6+ . For ortho series sum = 1.7 For para sum = 1.8

  21. At higher temperatures two become similar no/np= fo/fp=(2s+1)0/(2s+1)p= 3/1 ie 25% para and 75% ortho To calculate Cv, evaluate f for ortho and para Rotational heat capacity of ordinary H2is thus 3/4Cv(ortho)+ 1/4Cv(para)at a given temperature

  22. At low temperatures , H2molecules are in lowest possible rotational level r= J(J+1)h2/8 2I Since J=1 for ortho fr(ortho)= 9e-h2/4 2I For para J=o fr(para)=1e-0h2/8 2I=1 Cv = d/dT[RT2dlnZ/dT]v

  23. = d/dT[LkT2dlnf/dT] Since lnf = constant Cv(ortho) = 0 Cv(para) = 0 Rotational contribution to heat capacity is zero at low temperature As temperature is raised rotational contribution becomes significance and specific heat increases.

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