Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Question
Chapter 14, Problem 14.98E
Interpretation Introduction

Interpretation:

The energies of rotation for ammonia, NH3, as the rotational quantum number J ranges from 1 to 10 are to be determined. An energy level diagram for all the rotational levels is to be constructed. The degeneracies of the levels are to be determined.

Concept introduction:

Atoms of a molecule rotate in space about its moment of inertia. The rotational quantum number is represented by the symbol J. The z component of the rotational motion of the molecule is represented by MJ.

Expert Solution & Answer
Check Mark

Answer to Problem 14.98E

The energies of rotation for ammonia, NH3, as the rotational quantum number J ranges from 1 to 10 are determined below.

J K Erot/m1
1 -1 3753.185
1 0 1264.06
1 1 3753.185
2 -2 13748.68
2 -1 6281.305
2 0 3792.18
2 1 6281.305
2 2 13748.68
3 -3 29986.49
3 -2 17540.86
3 -1 10073.49
3 0 7584.36
3 1 10073.49
3 2 17540.86
3 3 29986.49
4 -4 52466.6
4 -3 35042.73
4 -2 22597.1
4 -1 15129.73
4 0 12640.6
4 1 15129.73
4 2 22597.1
4 3 52466.6
4 4 52466.6
5 -5 81189.03
5 -4 58786.9
5 -3 41363.03
5 -2 28917.4
5 -1 21450.03
5 0 18960.9
5 1 21450.03
5 2 28917.4
5 3 41363.03
5 4 58786.9
5 5 81189.03
6 -6 116153.8
6 -5 88773.39
6 -4 66371.26
6 -3 48947.39
6 -2 36501.76
6 -1 29034.39
6 0 26545.26
6 1 29034.39
6 2 36501.76
6 3 48947.39
6 4 66371.26
6 5 88773.39
6 6 116153.8
7 -7 157360.8
7 -6 125002.2
7 -5 97621.81
7 -4 75219.68
7 -3 57795.81
7 -2 45350.18
7 -1 37882.81
7 0 35393.68
7 1 37882.81
7 2 45350.18
7 3 57795.81
7 4 75219.68
7 5 97621.81
7 6 125002.2
7 7 157360.8
8 -8 204810.2
8 -7 167473.3
8 -6 135114.7
8 -5 107734.3
8 -4 85332.16
8 -3 67908.29
8 -2 55462.66
8 -1 47995.29
8 0 45506.16
8 1 47995.29
8 2 55462.66
8 3 67908.29
8 4 85332.16
8 5 107734.3
8 6 135114.7
8 7 167473.3
8 8 204810.2
9 -9 258501.8
9 -8 216186.7
9 -7 178849.8
9 -6 146491.2
9 -5 119110.8
9 -4 96708.7
9 -3 79284.83
9 -2 66839.2
9 -1 59371.83
9 0 56882.7
9 1 59371.83
9 2 66839.2
9 3 79284.83
9 4 96708.7
9 5 119110.8
9 6 146491.2
9 7 178849.8
9 8 216186.7
9 9 258501.8
10 -10 318435.8
10 -9 271142.4
10 -8 228827.3
10 -7 191490.4
10 -6 159131.8
10 -5 131751.4
10 -4 109349.3
10 -3 91925.43
10 -2 79479.8
10 -1 72012.43
10 0 69523.3
10 1 72012.43
10 2 79479.8
10 3 91925.43
10 4 109349.3
10 5 131751.4
10 6 159131.8
10 7 191490.4
10 8 228827.3
10 9 271142.4
10 10 318435.8

For the rotational quantum number J=1, the degeneracy is 3.

For the rotational quantum number J=2, the degeneracy is 5.

For the rotational quantum number J=3, the degeneracy is 7.

For the rotational quantum number J=4, the degeneracy is 9.

For the rotational quantum number J=5, the degeneracy is 11.

For the rotational quantum number J=6, the degeneracy is 13.

For the rotational quantum number J=7, the degeneracy is 15.

For the rotational quantum number J=8, the degeneracy is 17.

For the rotational quantum number J=9, the degeneracy is 19.

For the rotational quantum number J=10, the degeneracy is 21.

The energy level diagram for all the rotational levels is shown below.

Physical Chemistry, Chapter 14, Problem 14.98E , additional homework tip  1

Explanation of Solution

The formula to energy of rotation (Erot) is given by the formula below.

Erot=BJ(J+1)+(CB)K2 …(1)

Where,

J is the rotational quantum number.

K is the quantum number bounded by J.

The formula for B is given below.

B=h8π2Ibc …(2)

The formula for C is given below.

C=h8π2Icc …(3)

Where,

h is the Planck’s constant. (6.6×1034Js).

c is the speed of light. (3×108ms1).

The value of Ib is 4.413×1047kgm2.

Substitute the value of Ib, h and c in equation (2).

B=6.6×1034Js8×(3.14)2×4.413×1047kgm2×3×108ms1=632.03Jskg1m2m1s1×kgm2s-21J=632.03m1

The value of Ic is 2.806×1047kgm2.

Substitute the value of Ic, h and c in equation (3).

C=6.6×1034Js8×(3.14)2×2.806×1047kgm2×3×108ms1=3121.155Jskg1m2m1s1×kgm2s-21J=3121.155m1

The value of K is shown by the equation below.

K=Jto+J …(4)

The degeneracy is calculated by the formula given below.

Degeneracy=2J+1 …(5)

For the rotational quantum number J=1, the value of K is calculated below.

K=1to+1=1,0,1

The value of K is 1.

The value of J is 1.

Substitute the value of J in equation (5).

Degeneracy=2×1+1=3

Therefore, the degeneracy is 3.

Substitute the value of J, K, B and C in equation (1).

Erot=(632.03m1×1(1+1))+(3121.155m1632.03m1)(1)2=1264.06m1+2849.125m1=3753.185m1

Similarly the value of Erot for J=1 and corresponding values of K is given below.

J K Erot/m1
1 -1 3753.185
1 0 1264.06
1 1 3753.185

For the rotational quantum number J=2, the value of K is calculated below.

K=2to+2=2,1,0,1,2

Substitute the value of J in equation (5).

Degeneracy=2×2+1=5

Therefore, the degeneracy is 5.

Similarly the value of Erot for J=2 and corresponding values of K is given below.

J K Erot/m1
2 -2 13748.68
2 -1 6281.305
2 0 3792.18
2 1 6281.305
2 2 13748.68

For the rotational quantum number J=3, the value of K is calculated below.

K=3to+3=3,2,1,0,1,2,3

Substitute the value of J in equation (5).

Degeneracy=2×3+1=7

Therefore, the degeneracy is 7.

Similarly the value of Erot for J=3 and corresponding values of K is given below.

J K Erot/m1
3 -3 29986.49
3 -2 17540.86
3 -1 10073.49
3 0 7584.36
3 1 10073.49
3 2 17540.86
3 3 29986.49

For the rotational quantum number J=4, the value of K is calculated below.

K=4to+4=4,3,2,1,0,1,2,3,4

Substitute the value of J in equation (5).

Degeneracy=2×4+1=9

Therefore, the degeneracy is 9.

Similarly the value of Erot for J=4 and corresponding values of K is given below.

J K Erot/m1
4 -4 52466.6
4 -3 35042.73
4 -2 22597.1
4 -1 15129.73
4 0 12640.6
4 1 15129.73
4 2 22597.1
4 3 52466.6
4 4 52466.6

For the rotational quantum number J=5, the value of K is calculated below.

K=5to+5=5,4,3,2,1,0,1,2,3,4,5

Substitute the value of J in equation (5).

Degeneracy=2×5+1=11

Therefore, the degeneracy is 11.

Similarly the value of Erot for J=5 and corresponding values of K is given below.

J K Erot/m1
5 -5 81189.03
5 -4 58786.9
5 -3 41363.03
5 -2 28917.4
5 -1 21450.03
5 0 18960.9
5 1 21450.03
5 2 28917.4
5 3 41363.03
5 4 58786.9
5 5 81189.03

For the rotational quantum number J=6, the value of K is calculated below.

K=6to+6=6,5,4,3,2,1,0,1,2,3,4,5,6

Substitute the value of J in equation (5).

Degeneracy=2×6+1=13

Therefore, the degeneracy is 13.

Similarly the value of Erot for J=6 and corresponding values of K is given below.

J K Erot/m1
6 -6 116153.8
6 -5 88773.39
6 -4 66371.26
6 -3 48947.39
6 -2 36501.76
6 -1 29034.39
6 0 26545.26
6 1 29034.39
6 2 36501.76
6 3 48947.39
6 4 66371.26
6 5 88773.39
6 6 116153.8

For the rotational quantum number J=7, the value of K is calculated below.

K=7to+7=7,6,5,4,3,2,1,0,1,2,3,4,5,6,7

Substitute the value of J in equation (5).

Degeneracy=2×7+1=15

Therefore, the degeneracy is 15.

Similarly the value of Erot for J=7 and corresponding values of K is given below.

J K Erot/m1
7 -7 157360.8
7 -6 125002.2
7 -5 97621.81
7 -4 75219.68
7 -3 57795.81
7 -2 45350.18
7 -1 37882.81
7 0 35393.68
7 1 37882.81
7 2 45350.18
7 3 57795.81
7 4 75219.68
7 5 97621.81
7 6 125002.2
7 7 157360.8

For the rotational quantum number J=8, the value of K is calculated below.

K=8to+8=8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8

Substitute the value of J in equation (5).

Degeneracy=2×8+1=17

Therefore, the degeneracy is 17.

Similarly the value of Erot for J=8 and corresponding values of K is given below.

J K Erot/m1
8 -8 204810.2
8 -7 167473.3
8 -6 135114.7
8 -5 107734.3
8 -4 85332.16
8 -3 67908.29
8 -2 55462.66
8 -1 47995.29
8 0 45506.16
8 1 47995.29
8 2 55462.66
8 3 67908.29
8 4 85332.16
8 5 107734.3
8 6 135114.7
8 7 167473.3
8 8 204810.2

For the rotational quantum number J=9, the value of K is calculated below.

K=9to+9=9,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9

Substitute the value of J in equation (5).

Degeneracy=2×9+1=19

Therefore, the degeneracy is 19.

Similarly the value of Erot for J=9 and corresponding values of K is given below.

J K Erot
9 -9 258501.8
9 -8 216186.7
9 -7 178849.8
9 -6 146491.2
9 -5 119110.8
9 -4 96708.7
9 -3 79284.83
9 -2 66839.2
9 -1 59371.83
9 0 56882.7
9 1 59371.83
9 2 66839.2
9 3 79284.83
9 4 96708.7
9 5 119110.8
9 6 146491.2
9 7 178849.8
9 8 216186.7
9 9 258501.8

For the rotational quantum number J=10, the value of K is calculated below.

K=10to+10=10,9,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9,10

Substitute the value of J in equation (5).

Degeneracy=2×10+1=21

Therefore, the degeneracy is 21.

Similarly the value of Erot for J=10 and corresponding values of K is given below.

J K Erot/m1
10 -10 318435.8
10 -9 271142.4
10 -8 228827.3
10 -7 191490.4
10 -6 159131.8
10 -5 131751.4
10 -4 109349.3
10 -3 91925.43
10 -2 79479.8
10 -1 72012.43
10 0 69523.3
10 1 72012.43
10 2 79479.8
10 3 91925.43
10 4 109349.3
10 5 131751.4
10 6 159131.8
10 7 191490.4
10 8 228827.3
10 9 271142.4
10 10 318435.8

The energy level diagram for all the rotational levels is shown below.

Physical Chemistry, Chapter 14, Problem 14.98E , additional homework tip  2

Figure 1

Conclusion

The energies of rotation for ammonia, NH3, as the rotational quantum number J ranges from 1 to 10 have been rightfully stated. An energy level diagram for all the rotational levels has been rightfully constructed. The degenracies of the levels are rightfully stated.

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