Question:
Rotational spectrum of a diatomic molecule consists of lines of equal spacing with an interval of \(20.0\text{ cm}^{-1}\). Its moment of inertia is found to be \(I_0 \times 10^{-47}\text{ kg.m}^2\), the value of \(I_0\) (rounded off to one decimal place) is ______.
(\(h = 6.6 \times 10^{-34}\text{ J.s}\), speed of light in vacuum \(c = 3 \times 10^8\text{ m.s}^{-1}\))
Rotational spectrum of a diatomic molecule consists of lines of equal spacing with an interval of \(20.0\text{ cm}^{-1}\). Its moment of inertia is found to be \(I_0 \times 10^{-47}\text{ kg.m}^2\), the value of \(I_0\) (rounded off to one decimal place) is ______.
(\(h = 6.6 \times 10^{-34}\text{ J.s}\), speed of light in vacuum \(c = 3 \times 10^8\text{ m.s}^{-1}\))
(\(h = 6.6 \times 10^{-34}\text{ J.s}\), speed of light in vacuum \(c = 3 \times 10^8\text{ m.s}^{-1}\))
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Hint:
The spacing between adjacent rotational lines equals \(2B_e\), and \(B_e = h/(8\pi^2 c I)\). Use \(c\) in cm/s since the spacing is given in \(\text{cm}^{-1}\).
The spacing between adjacent rotational lines equals \(2B_e\), and \(B_e = h/(8\pi^2 c I)\). Use \(c\) in cm/s since the spacing is given in \(\text{cm}^{-1}\).
Updated On: Jul 28, 2026
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Correct Answer: 2.8
Solution and Explanation
Step 1: Understanding the Concept:
A diatomic molecule behaves like a rigid rotor. Its allowed rotational energy levels are \(E_J = B_e \, hc \, J(J+1)\), where \(J = 0, 1, 2, ...\) is the rotational quantum number and \(B_e\) (in \(\text{cm}^{-1}\)) is the rotational constant. Transitions follow the selection rule \(\Delta J = \pm 1\).
Step 2: Key Formula or Approach:
For a transition from \(J\) to \(J+1\), the line appears at wavenumber \(\tilde{\nu} = 2B_e(J+1)\). Consecutive lines therefore differ by a constant spacing:
\[ \Delta\tilde{\nu} = 2B_e \]
The rotational constant is related to the moment of inertia \(I\) by:
\[ B_e = \frac{h}{8\pi^2 c I} \]
Step 3: Detailed Explanation:
The line spacing is given as \(\Delta\tilde{\nu} = 20.0\text{ cm}^{-1}\), so:
\[ B_e = \frac{\Delta\tilde{\nu}}{2} = 10.0\text{ cm}^{-1} \]
Rearrange the formula for \(B_e\) to solve for \(I\), remembering to use \(c\) in \(\text{cm/s}\) since \(B_e\) is in \(\text{cm}^{-1}\): \(c = 3\times 10^8\text{ m/s} = 3\times 10^{10}\text{ cm/s}\).
\[ I = \frac{h}{8\pi^2 c B_e} = \frac{6.6\times 10^{-34}}{8\pi^2 \times (3\times 10^{10}) \times 10.0} \]
\[ I = \frac{6.6\times 10^{-34}}{2.369\times 10^{13}} = 2.79\times 10^{-47}\text{ kg.m}^2 \]
Final Answer:
So the moment of inertia is \(2.79\times 10^{-47}\text{ kg.m}^2\), which rounds to \(I_0 = 2.8\). \[ \boxed{I_0 = 2.8} \]
A diatomic molecule behaves like a rigid rotor. Its allowed rotational energy levels are \(E_J = B_e \, hc \, J(J+1)\), where \(J = 0, 1, 2, ...\) is the rotational quantum number and \(B_e\) (in \(\text{cm}^{-1}\)) is the rotational constant. Transitions follow the selection rule \(\Delta J = \pm 1\).
Step 2: Key Formula or Approach:
For a transition from \(J\) to \(J+1\), the line appears at wavenumber \(\tilde{\nu} = 2B_e(J+1)\). Consecutive lines therefore differ by a constant spacing:
\[ \Delta\tilde{\nu} = 2B_e \]
The rotational constant is related to the moment of inertia \(I\) by:
\[ B_e = \frac{h}{8\pi^2 c I} \]
Step 3: Detailed Explanation:
The line spacing is given as \(\Delta\tilde{\nu} = 20.0\text{ cm}^{-1}\), so:
\[ B_e = \frac{\Delta\tilde{\nu}}{2} = 10.0\text{ cm}^{-1} \]
Rearrange the formula for \(B_e\) to solve for \(I\), remembering to use \(c\) in \(\text{cm/s}\) since \(B_e\) is in \(\text{cm}^{-1}\): \(c = 3\times 10^8\text{ m/s} = 3\times 10^{10}\text{ cm/s}\).
\[ I = \frac{h}{8\pi^2 c B_e} = \frac{6.6\times 10^{-34}}{8\pi^2 \times (3\times 10^{10}) \times 10.0} \]
\[ I = \frac{6.6\times 10^{-34}}{2.369\times 10^{13}} = 2.79\times 10^{-47}\text{ kg.m}^2 \]
Final Answer:
So the moment of inertia is \(2.79\times 10^{-47}\text{ kg.m}^2\), which rounds to \(I_0 = 2.8\). \[ \boxed{I_0 = 2.8} \]
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