Question

What are the moment of inertia of ${}^1{\mathbf{H}}^{19}\mathbf{F}$ and${}^1{\mathbf{H}}^{81}\mathbf{Br}$ , expressed in kg m² ? Compute the spacing $\mathbf{\mu }\mathbf{=}\mathbf{\Delta E}\mathbf{/}\mathbf{h}$ of the rotational states, in s^-1 , between $\mathbf{J}\mathbf{=}\mathbf{0}$ and $\mathbf{1}$ and between $\mathbf{J}\mathbf{=}\mathbf{1}$ anddata-custom-editor="chemistry" $\mathbf{2}$. Explain, in one sentence, why the large change in mass from 19 to 81 causes only a small change in rotational energy differences.

Short answer

Moment of inertia of ${}^1{H}^{19}F$is $0.95\times {10}^{-3}kg$.

Moment of inertia of ${}^1{H}^{81}Br$is $9.9\times {10}^{-4}kg$.

Due to equal spacing in hetero atomic molecules, there’s only a small change with increase in atomic numbers.

Step-by-step solution

Moment of inertia

The moment of inertia is the amount of torque required to rotate a molecule. The molecular entropy is calculated using the product of the three moments of inertia.

Given information and formula used

• Relative atomic mass of ${}^{\mathbf{1}}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{u}$
• Relative atomic mass of ${}^{\mathbf{19}}\mathbf{F}\mathbf{=}\mathbf{19}\mathbf{u}$
• Relative atomic mass of ${}^{\mathbf{81}}\mathbf{Br}\mathbf{=}\mathbf{81}\mathbf{u}$

Reduced mass of body is given by:

$\mathbf{\mu }\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}$

Moment of inertia of a molecule is given by: $\mathbf{I}\mathbf{=}{\mathbf{\mu R}}_{\mathbf{e}}^{\mathbf{2}}$

Rotational constant of a molecule is given by:

$\stackrel{\mathbf{~}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{h}}{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{Ic}}$

Frequency change is given by: $\mathbf{\nu }\mathbf{=}\mathbf{B}\mathbf{\left[}{\mathbf{J}}_{\mathbf{f}}\mathbf{(}{\mathbf{J}}_{\mathbf{f}}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}{\mathbf{J}}_{\mathbf{i}}\mathbf{(}{\mathbf{J}}_{\mathbf{i}}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\right]}$

Energy is given by: $\mathbf{\Delta {\rm E}}\mathbf{=}\mathbf{h\nu }$

Calculation of reduced mass

Calculate the reduced mass of ${}^1{H}^{19}F$ by substituting the values in the formula.

$\begin{array}{l}\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ =\frac{\left(1\right)\left(19\right)}{1+19}\\ =0.95\times {10}^{-3}\mathrm{kg}\end{array}$

Calculate the reduced mass of ${}^1{H}^{81}Br$ by substituting the values in the formula.

$\begin{array}{l}\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ =\frac{\left(1\right)\left(81\right)}{1+81}\\ =9.9\times {10}^{-4}\mathrm{kg}\end{array}$

Calculation of moment of inertia

Calculate moment of inertia for ${}^1{H}^{19}F$.

$\begin{array}{c}\mathrm{I}={\mathrm{\mu R}}_{\mathrm{e}}^2\\ =(0.95\times {10}^{-3}){(0.092\times {10}^{-9})}^2\\ =8.04\times {10}^{-{24}^{{}^{}}}\mathrm{kg}{\mathrm{m}}^2\end{array}$

Calculate moment of inertia for ${}^1{H}^{81}Br$.

$\begin{array}{c}\mathrm{I}={\mathrm{\mu R}}_{\mathrm{e}}^2\\ =(9.9\times {10}^{-4}){(0.141\times {10}^{-9})}^2\\ =1.95\times {10}^{-{23}^{{}^{}}}\mathrm{kg}{\mathrm{m}}^2\end{array}$

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Calculation of rotational constant 

Calculate the rotational constant of${}^1{H}^{19}F$ by substituting the values in the formula.

$\begin{array}{c}\tilde{\mathrm{B}}=\frac{\mathrm{h}}{8{\mathrm{\pi }}^2\mathrm{Ic}}\\ =\frac{6.626\times {10}^{-34}}{8{\mathrm{\pi }}^2(8.04\times {10}^{-24})}\\ =1.04\times {10}^{-12}{\mathrm{s}}^{-1}\end{array}$

Calculate the rotational constant of${}^1{H}^{81}Br$by substituting the values in the formula.

$\begin{array}{c}\tilde{\mathrm{B}}=\frac{\mathrm{h}}{8{\mathrm{\pi }}^2\mathrm{Ic}}\\ =\frac{6.626\times {10}^{-34}}{8{\mathrm{\pi }}^2(1.93\times {10}^{-23})}\\ =4.30\times {10}^{-13}{\mathrm{s}}^{-1}\end{array}$

Calculation of energy level of  1H19F

Calculate the frequency for transition $J=0\to J=1$.

$\begin{array}{c}\mathrm{\nu }=\mathrm{B}\left[{\mathrm{J}}_{\mathrm{f}}({\mathrm{J}}_{\mathrm{f}}+1)-{\mathrm{J}}_{\mathrm{i}}({\mathrm{J}}_{\mathrm{i}}+1)\right]\\ =(1.04\times {10}^{-12})\left[1(1+1)-0(0_{\mathrm{i}}+1)\right]\\ =1.04\times {10}^{-12}\times 2\\ =2.08\times {10}^{-12}{\mathrm{s}}^{-1}\end{array}$

Put the values and find energy for transition $J=0\to J=1$

$\begin{array}{c}\mathrm{\Delta {\rm E}}=\mathrm{h\nu }\\ \mathrm{\Delta {\rm E}}=(6.626\times {10}^{-34})(2.08\times {10}^{-12})\\ \mathrm{\Delta {\rm E}}=13.78\times {10}^{-46}\mathrm{J}\\ (\because 1\mathrm{J}=5.034\times {10}^{22}{\mathrm{cm}}^{-1})\\ \mathrm{\Delta {\rm E}}=69.36\times {10}^{-24}{\mathrm{cm}}^{-1}\end{array}$

Now, calculate the frequency for transition $J=1\to J=2$.

$\begin{array}{c}\mathrm{\nu }=\mathrm{B}\left[{\mathrm{J}}_{\mathrm{f}}({\mathrm{J}}_{\mathrm{f}}+1)-{\mathrm{J}}_{\mathrm{i}}({\mathrm{J}}_{\mathrm{i}}+1)\right]\\ =(1.04\times {10}^{-12})\left[2(2+1)-1(1+1)\right]\\ =1.04\times {10}^{-12}\times 6\\ =6.24\times {10}^{-12}{\mathrm{s}}^{-1}\end{array}$

Put the values and find energy for transition $J=1\to J=2$.

$\begin{array}{l}\mathrm{\Delta {\rm E}}=\mathrm{h\nu }\\ \mathrm{\Delta {\rm E}}=(6.626\times {10}^{-34})(6.24\times {10}^{-12})\\ \mathrm{\Delta {\rm E}}=41.34\times {10}^{-46}\mathrm{J}\\ (\because 1\mathrm{J}=5.034\times {10}^{22}{\mathrm{cm}}^{-1})\\ \mathrm{\Delta {\rm E}}=208.10\times {10}^{-24}{\mathrm{cm}}^{-1}\end{array}$

Calculation of energy level of  1H81Br

Calculate the frequency for transition $J=0\to J=1$.

$\begin{array}{c}\mathrm{\nu }=\mathrm{B}\left[{\mathrm{J}}_{\mathrm{f}}({\mathrm{J}}_{\mathrm{f}}+1)-{\mathrm{J}}_{\mathrm{i}}({\mathrm{J}}_{\mathrm{i}}+1)\right]\\ =(4.30\times {10}^{-13})\left[1(1+1)-0(0_{\mathrm{i}}+1)\right]\\ =4.30\times {10}^{-13}\times 2\\ =8.6\times {10}^{-13}{\mathrm{s}}^{-1}\end{array}$

Put the values and find energy for transition$J=0\to J=1$ .

$\begin{array}{l}\mathrm{\Delta {\rm E}}=\mathrm{h\nu }\\ \mathrm{\Delta {\rm E}}=(6.626\times {10}^{-34})(8.6\times {10}^{-13})\\ \mathrm{\Delta {\rm E}}=59.98\times {10}^{-47}\mathrm{J}\\ (\because 1\mathrm{J}=5.034\times {10}^{22}{\mathrm{cm}}^{-1})\\ \mathrm{\Delta {\rm E}}=286.83\times {10}^{-25}{\mathrm{cm}}^{-1}\end{array}$

Now, calculate the frequency for transition $J=1\to J=2$.

$\begin{array}{c}\nu =B\left[J_f(J_f+1)-J_i(J_i+1)\right]\\ =(4.30\times {10}^{-13})\left[2(2+1)-1(1+1)\right]\\ =4.30\times {10}^{-13}\times 6\\ =25.8\times {10}^{-13}{s}^{-1}\end{array}$

Put the values and find energy for transition $J=1\to J=2$

$\begin{array}{l}\mathrm{\Delta {\rm E}}=\mathrm{h\nu }\\ \mathrm{\Delta {\rm E}}=(6.626\times {10}^{-34})(25.8\times {10}^{-13})\\ \mathrm{\Delta {\rm E}}=170.95\times {10}^{-47}\mathrm{J}\\ (\because 1\mathrm{J}=5.034\times {10}^{22}{\mathrm{cm}}^{-1})\\ \mathrm{\Delta {\rm E}}=860.56\times {10}^{-25}{\mathrm{cm}}^{-1}\end{array}$

Rotational energy difference

Due to equally spaced line separated by $2\tilde{B}$ in the spectrum of hetro nuclear diatomic molecule, there is a small change in rotational energy difference while increasing atomic mass from 19 to 81.