Molecular vibration: A certain diatomic molecule emits a photon of energy 1.20 eV when it makes a transition from the n = 1 vibrational state to the next lower vibrational state. What is the frequency of vibration of the molecule? (h = 6.626 × 10^-34 J ∙ s, ħ = 1.055 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 1.93 × 10¹⁴ Hz B) 2.90 × 10¹⁴ Hz C) 4.35 × 10¹⁴ Hz D) 1.82 × 10¹⁵ Hz E) 2.73 × 10¹⁵ Hz

Question

Quizplus · Accepted Answer

First, we need to find the energy difference between the two vibrational levels:

ΔE = E(n=1) - E(n=0) = 1.20 eV

We can convert this to joules:

ΔE = (1.20 eV)(1.60 × 10^-19 J/eV) = 1.92 × 10^-19 J

Since energy is related to frequency and Planck's constant by the equation:

E = hν

We can rearrange this to solve for frequency:

ν = E/h

Plugging in the values we have:

ν = (1.92 × 10^-19 J)/(6.626 × 10^-34 J∙s)

ν = 2.90 × 10^14 Hz

This matches choice B, so the answer is B.

Molecular Vibration: a Certain Diatomic Molecule Emits a Photon of Energy