Question

The rotational constant for ${}^{\mathbf{14}}\mathbf{N}_{\mathbf{2}}$, measured for the first time using rotational Raman spectroscopy (see Fig. 20.9) is role="math" localid="1663691379622" $\mathbf{1}\mathbf{.}\mathbf{99}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$. Calculate (a) the moment of inertia I of in $\mathbf{kg}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$ , (b) the energies of the first three excited rotational levels, relative to that of the ground level, in wave numbers, and (c) the frequency of the $\mathbf{J}\mathbf{=}\mathbf{0}\mathbf{\to }\mathbf{J}\mathbf{=}\mathbf{1}$ transition in GHz.

Short answer

(a)The moment of inertia is for${}^{14}{\mathrm{N}}_2$ is role="math" localid="1663691652471" $1.407\times {10}^{-44 } \mathrm{kg} {\mathrm{m}}^2$.

(b)The energy of first excited state is $3.98{\mathrm{cm}}^{-1}$.

The energy of second excited state is $7.96{\mathrm{cm}}^{-1}$.

The energy of third excited state is $11.94{\mathrm{cm}}^{-1}$.

(c) The frequency of the vibration is $1.19\times {10}^2\mathrm{GHz}$.

Step-by-step solution

Moment of Inertia and Rotational Spectroscopy

The total of the product of each atom’s mass multiplied by the square of its distance (r) from the rotating axis through the molecule’s center of mass (m) is the molecule’s moment of inertia (I) connected with an axis.

The energy measurement of transitions between “molecule’s rotational states” (quantized) in the gas phase is the subject of rotational spectroscopy.

Given information and formula used

Rotational constant for${}^{\mathbf{14}}{\mathbf{N}}_{\mathbf{2}}$is $1.99{\mathrm{cm}}^{-1}$.

Moment of inertia (l) of a molecule is given by:

$\mathbf{I}\mathbf{=}\frac{\mathbf{h}}{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}\stackrel{\mathbf{~}}{\mathbf{B}}\mathbf{c}}$….. (i)

where$\stackrel{\mathbf{~}}{\mathbf{B}}$is rotational constant.

Energy is given by the formula:$\mathbf{\Delta E}\mathbf{=}\stackrel{\mathbf{~}}{\mathbf{B}}\left[J_f(J_f+1)-J_i(J_i+1)\right]$ ….. (ii)

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

Calculation of moment of inertia

a) Substitute the values in equation (i).

$$\begin{gathered} \mathrm{I}=\frac{6.63\times {10}^{-34}}{8\times {(3.14)}^2\times 1.99\times 3\times {10}^8} \\ \mathrm{I}=1.407\times {10}^{-44 } \mathrm{kg} {\mathrm{m}}^2 \end{gathered}$$

Thus, the moment of inertia is for is $1.407\times {10}^{-44 } \mathrm{kg} {\mathrm{m}}^2$.

Calculation of energies of excited states

b) For first excited state, substitute ${\mathrm{J}}_{\mathrm{f}}=1 \mathrm{and} {\mathrm{J}}_{\mathrm{i}}=0$in the given equation (ii).

The known values are substituted.

$$\begin{gathered} \mathrm{\Delta E}=1.99\left\{\right(1\left)\right(1+1)-(0\left)\right(0+1\left)\right\} \\ =1.99\times 2 \\ =3.98{\mathrm{cm}}^{-1} \end{gathered}$$

Thus, the energy of the first excited state is $3.98{\mathrm{cm}}^{-1}$.

For the second excited state, substitute${\mathrm{J}}_{\mathrm{f}}=2 \mathrm{and} {\mathrm{J}}_{\mathrm{i}}=1$in the given equation (ii).

$$\begin{gathered} \Delta E=1.99\left\{\right(2\left)\right(2+1)-(1\left)\right(1+1\left)\right\} \\ =1.99\left\{\right(2\left)\right(3)-(1\left)\right(2\left)\right\} \\ =1.99(6-2) \\ =1.99\times 4 \\ =7.96c{m}^{-1} \end{gathered}$$

Thus, the energy of second excited state is$7.96{\mathrm{cm}}^{-1}$.

For the third excited state, substitute${\mathrm{J}}_{\mathrm{f}}=3 \mathrm{and} {\mathrm{J}}_{\mathrm{i}}=2$in the given equation (ii).

$$\begin{gathered} \Delta E=1.99\left\{\right(3\left)\right(3+1)-(2\left)\right(2+1\left)\right\} \\ =1.99\left\{\right(3\left)\right(4)-(2\left)\right(3\left)\right\} \\ =1.99(12-6) \\ =1.99\times 6 \\ =11.94c{m}^{-1} \end{gathered}$$

Thus, the energy of the third excited state is$11.94{\mathrm{cm}}^{-1}$.

Calculation of frequency of transition state

c) Substitute the values in given equation for the transition from J = 0 → J = 1.

$\begin{array}{rcl}\mathrm{\Delta E}& =& \mathrm{h\nu }\\ \mathrm{\nu }& =& \frac{\mathrm{\Delta E}}{\mathrm{h}}\end{array}$

The known values are substituted.

$$\begin{gathered} \nu =\frac{3.98\times 1.989\times {10}^{-23}}{6.63\times {10}^{-34}} \\ (\because {\text{1cm}}^{\text{- 1}}=1.986\times {10}^{-23}J) \\ =1.192\times {10}^{11} {\text{s}}^{\text{- 1}} \\ (\because \text{1} \text{GHz}={10}^9{s}^{-1}) \\ =1.19\times {10}^2\text{GHz} \end{gathered}$$

Thus, the frequency of the vibration is $1.19\times {10}^2\mathrm{GHz}$.