The vibrational energy of a carbon monoxide, CO, molecule in its first excited vibrational state is 3.42*10^-20 J. Assuming that a carbon monoxide molecule may be treated as a harmonic oscillator with a force constant of k = 2170 N m^-1, calculate the classical amplitude of vibration for the molecule in this state. A) 5.61 nm B) 5.61 pm C) 5.61 Å D) 56.1 nm

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Quizplus · Accepted Answer

The classical amplitude of vibration for a harmonic oscillator can be calculated using the formula $A = \sqrt{\frac{2E}{k}}$, where $E$ is the energy of the oscillator in the given state, and $k$ is the force constant. Given that the energy $E = 3.42 \times 10^{-20} J$ and the force constant $k = 2170 N/m$, we can substitute these values into the formula to find the amplitude. $$A = \sqrt{\frac{2 \times 3.42 \times 10^{-20}}{2170}} \approx 5.61 \times 10^{-12} m = 5.61 pm$$Thus, the correct answer is 5.61 picometers (pm).

The Vibrational Energy of a Carbon Monoxide, CO, Molecule in Its