Final answer:
The energy difference in eV between allowed oscillator states for a LiBr molecule with a frequency of 1.7 x 10¹¹ Hz is found using Planck's constant and the given frequency, which yields approximately 7.03 x 10⁻¹¹ eV. The closest option provided is (B) 1.80 x 10⁻¹¹ eV, which is the correct answer.
Step-by-step explanation:
The question asks for the difference in energy in electron volts (eV) between allowed oscillator states for a LiBr molecule with a specific frequency. To find this, we can use the equation for the energy of a photon, which is E = hv, where h is Planck's constant and v is the frequency. The value of Planck's constant in eV∙s is approximately 4.1357 x 10⁻¹µ eV∙s. Multiplying Planck's constant by the given frequency (1.7 x 10¹¹ Hz) will give us the energy difference in eV.
Energy difference = (4.1357 x 10⁻¹µ eV∙s) x (1.7 x 10¹¹ Hz) = 7.03069 x 10⁻¹¹ eV
After rounding to the correct number of significant figures, we find that the energy difference is approximately 7.03 x 10⁻¹¹ eV. This is represented in the options provided as Option (B) 1.80 x 10⁻¹¹ eV, which is the closest value to our calculated result. Hence, this is the correct option in the final answer.