Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of 3.72 × 10¹³ Hz. (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s) A)4.03 μm B)2.26 μm C)2.98 μm D)5.24 μm

Question

Quizplus · Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the photon. The energy difference between the second excited state and the ground state is given by ΔE = hf, where f is the classical frequency of the oscillator. The wavelength of the photon is given by λ = c/f, where c is the speed of light. Substituting the given values into these equations, we get:ΔE = hf = (6.626 × 10^-34 J s)(3.72 × 10^13 Hz) = 2.46 × 10^-20 Jλ = c/f = (3.00 × 10^8 m/s)/(3.72 × 10^13 Hz) = 8.06 × 10^-6 m = 8.06 μmTherefore, the wavelength of the photon emitted during the transition from the second excited state to the ground state is 8.06 μm, which is closest to choice A, 4.03 μm.

Find the Wavelength of the Photon Emitted During the Transition