Chapter 14, Problem 14.98E

Interpretation Introduction

Interpretation:

The energies of rotation for ammonia, $\text{NH}{\text{3}}_{}$, 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 $M_J$.

Answer to Problem 14.98E

The energies of rotation for ammonia, $\text{NH}{\text{3}}_{}$, as the rotational quantum number $J$ ranges from $1$ to $10$ are determined below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

Explanation of Solution

The formula to energy of rotation ($E_{\text{rot}}$) is given by the formula below.

$E_{\text{rot}}=BJ\left(J+1\right)+\left(C-B\right)K^2$ …(1)

Where,

• $J$ is the rotational quantum number.

• $K$ is the quantum number bounded by $J$.

The formula for $B$ is given below.

$B=\frac{h}{8{\pi }^2{I}_{\text{b}}c}$ …(2)

The formula for $C$ is given below.

$C=\frac{h}{8{\pi }^2{I}_{\text{c}}c}$ …(3)

Where,

• $h$ is the Planck’s constant. $\left(6.6\times {10}^{-34}\text{J}\text{s}\right)$.

• $c$ is the speed of light. $\left(3\times {10}^8\text{m}{\text{s}}^{-1}\right)$.

The value of $I_{\text{b}}$ is $4.413\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}$.

Substitute the value of $I_{\text{b}}$, $h$ and $c$ in equation (2).

$\begin{array}{rll}B& =& \frac{6.6\times {10}^{-34}\text{J}\text{s}}{8\times {\left(3.14\right)}^2\times 4.413\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\\ & =& 632.03\text{J}\text{s}{\text{kg}}^{-1}{\text{m}}^{-\text{2}}{\text{m}}^{-1}{\text{s}}^1\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}{\text{1}\text{J}}\\ & =& 632.03{\text{m}}^{-1}\end{array}$

The value of $I_{\text{c}}$ is $2.806\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}$.

Substitute the value of $I_{\text{c}}$, $h$ and $c$ in equation (3).

$\begin{array}{rll}C& =& \frac{6.6\times {10}^{-34}\text{J}\text{s}}{8\times {\left(3.14\right)}^2\times 2.806\times {10}^{-47}\text{kg}{\text{m}}^{\text{2}}\times 3\times {10}^8\text{m}{\text{s}}^{-1}}\\ & =& 3121.155\text{J}\text{s}{\text{kg}}^{-1}{\text{m}}^{-\text{2}}{\text{m}}^{-1}{\text{s}}^1\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}{\text{1}\text{J}}\\ & =& 3121.155{\text{m}}^{-1}\end{array}$

The value of $K$ is shown by the equation below.

$K=-J\text{to}+J$ …(4)

The degeneracy is calculated by the formula given below.

$\text{Degeneracy}=2J+1$ …(5)

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

$\begin{array}{rll}K& =& -1\text{to}+1\\ & =& -1,0,1\end{array}$

The value of $K$ is $-1$.

The value of $J$ is $1$.

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 1+1\\ & =& 3\end{array}$

Therefore, the degeneracy is $3$.

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

$\begin{array}{rll}{E}_{\text{rot}}& =& \left(632.03{\text{m}}^{-1}\times 1\left(1+1\right)\right)+\left(3121.155{\text{m}}^{-1}-632.03{\text{m}}^{-1}\right){\left(-1\right)}^2\\ & =& 1264.06{\text{m}}^{-1}+2849.125{\text{m}}^{-1}\\ & =& 3753.185{\text{m}}^{-1}\end{array}$

Similarly the value of $E_{\text{rot}}$ for $J=1$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -2\text{to}+2\\ & =& -2,-1,0,1,2\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 2+1\\ & =& 5\end{array}$

Therefore, the degeneracy is $5$.

Similarly the value of $E_{\text{rot}}$ for $J=2$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -3\text{to}+3\\ & =& -3,-2,-1,0,1,2,3\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 3+1\\ & =& 7\end{array}$

Therefore, the degeneracy is $7$.

Similarly the value of $E_{\text{rot}}$ for $J=3$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -4\text{to}+4\\ & =& -4,-3,-2,-1,0,1,2,3,4\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 4+1\\ & =& 9\end{array}$

Therefore, the degeneracy is $9$.

Similarly the value of $E_{\text{rot}}$ for $J=4$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -5\text{to}+5\\ & =& -5,-4,-3,-2,-1,0,1,2,3,4,5\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 5+1\\ & =& 11\end{array}$

Therefore, the degeneracy is $11$.

Similarly the value of $E_{\text{rot}}$ for $J=5$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -6\text{to}+6\\ & =& -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 6+1\\ & =& 13\end{array}$

Therefore, the degeneracy is $13$.

Similarly the value of $E_{\text{rot}}$ for $J=6$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -7\text{to}+7\\ & =& -7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 7+1\\ & =& 15\end{array}$

Therefore, the degeneracy is $15$.

Similarly the value of $E_{\text{rot}}$ for $J=7$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -8\text{to}+8\\ & =& -8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 8+1\\ & =& 17\end{array}$

Therefore, the degeneracy is $17$.

Similarly the value of $E_{\text{rot}}$ for $J=8$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

$\begin{array}{rll}K& =& -9\text{to}+9\\ & =& -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 9+1\\ & =& 19\end{array}$

Therefore, the degeneracy is $19$.

Similarly the value of $E_{\text{rot}}$ for $J=9$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}$
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.

$\begin{array}{rll}K& =& -10\text{to}+10\\ & =& -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10\end{array}$

Substitute the value of $J$ in equation (5).

$\begin{array}{rll}\text{Degeneracy}& =& 2\times 10+1\\ & =& 21\end{array}$

Therefore, the degeneracy is $21$.

Similarly the value of $E_{\text{rot}}$ for $J=10$ and corresponding values of $K$ is given below.

$J$	$K$	$E_{\text{rot}}/{\text{m}}^{-1}$
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.

Figure 1

Conclusion

The energies of rotation for ammonia, $\text{NH}{\text{3}}_{}$, 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.