When a hypothetical diamotic molecule having atoms 0.8890 nm apart undergoes a rotational transition from the l=2 state to the next lower state, it gives up a photon having energy 8.850 * 10^-4 eV. When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up 0.2540 eV. Find the force constant of this molecule.

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### Educational Text on Rotational and Vibrational Transitions in Diatomic Molecules

**Problem Statement:**
When a hypothetical diatomic molecule having atoms 0.8890 nm apart undergoes a rotational transition from the \( l = 2 \) state to the next lower state, it gives up a photon having energy \( 8.850 \times 10^{-4} \) eV. When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up \( 0.2540 \) eV. Find the force constant of this molecule.

**Analysis:**

1. **Rotational Transition:** 
   - Given:
     - Distance between atoms \( r = 0.8890 \) nm = \( 0.8890 \times 10^{-9} \) m
     - Transition from \( l = 2 \) to \( l = 1 \)
     - Energy of photon emitted \( \Delta E_{\text{rot}} = 8.850 \times 10^{-4} \) eV
   
   Rotational energy levels for a diatomic molecule are given by:
   \[
   E_l = \frac{l(l + 1)h^2}{8\pi^2I}
   \]
   where \( I \) (moment of inertia) is \( \mu r^2 \) and \( \mu \) is the reduced mass of the system.

2. **Vibrational Transition:** 
   - Given:
     - Energy difference \( \Delta E_{\text{vib}} = 0.2540 \) eV
  
   Vibrational energy levels for a diatomic molecule are given by:
   \[
   E_v = \left( v + \frac{1}{2} \right) h \nu
   \]
   where \( \nu \) is the vibrational frequency and \( v \) is the vibrational quantum number.

**Objective:**
To find the force constant \( k \) of this molecule, which is related to the vibrational frequency \( \nu \) by:
\[
\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}
\]
By solving the given problem, we can establish the relationship and find the desired force constant \( k \).

**Steps:**

1. Calculate the moment of inertia \( I \
Transcribed Image Text:### Educational Text on Rotational and Vibrational Transitions in Diatomic Molecules **Problem Statement:** When a hypothetical diatomic molecule having atoms 0.8890 nm apart undergoes a rotational transition from the \( l = 2 \) state to the next lower state, it gives up a photon having energy \( 8.850 \times 10^{-4} \) eV. When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up \( 0.2540 \) eV. Find the force constant of this molecule. **Analysis:** 1. **Rotational Transition:** - Given: - Distance between atoms \( r = 0.8890 \) nm = \( 0.8890 \times 10^{-9} \) m - Transition from \( l = 2 \) to \( l = 1 \) - Energy of photon emitted \( \Delta E_{\text{rot}} = 8.850 \times 10^{-4} \) eV Rotational energy levels for a diatomic molecule are given by: \[ E_l = \frac{l(l + 1)h^2}{8\pi^2I} \] where \( I \) (moment of inertia) is \( \mu r^2 \) and \( \mu \) is the reduced mass of the system. 2. **Vibrational Transition:** - Given: - Energy difference \( \Delta E_{\text{vib}} = 0.2540 \) eV Vibrational energy levels for a diatomic molecule are given by: \[ E_v = \left( v + \frac{1}{2} \right) h \nu \] where \( \nu \) is the vibrational frequency and \( v \) is the vibrational quantum number. **Objective:** To find the force constant \( k \) of this molecule, which is related to the vibrational frequency \( \nu \) by: \[ \nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \] By solving the given problem, we can establish the relationship and find the desired force constant \( k \). **Steps:** 1. Calculate the moment of inertia \( I \
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