Question

The cesium iodide (CsI) molecule has an atomic separation of 0.127 nm.

(a) Determine the energy of the first excited rotational state, with J = 1. (Anser in meV)

(b) Find the frequency of the photon absorbed in the J = 0 to J = 1 transition. (Answer in GHz)

Answer 1

The effective mass of the Cesium Idoide (CsI) molecule (μ) = m₁m₂ / (m₁ + m₂)

Therefore, μ = (132.9 * 126.9) / (132.9 + 126.9) = 64.92 amu = 64.92 * 1.66 * 10^-27 kg = 1.078 * 10^-25 kg

Moment of Inertia (I) = μ * R² = 1.078 * 10^-25 * 0.127 * 10^-9 * 0.127 * 10^-9 = 1.738 * 10^-45 kg-m²

(a)

We know, E = 0.5 * I * ω² = (Iω)² / 2I = J(J + 1)h^'2 / 2I

Energy of the first excited rotational state with J = 1 =>

E = h^'2 / I = h² / 4 * pi² * I = (6.626 * 10^-34)² / (4 * pi² * 1.738 * 10^-45) = 6.398 * 10^-24 J = 6.398 * 10^-24 * 6.241509 * 10²¹ meV = 39.94 * 10^-3 meV

Therefore, energy of first excieted rotational state = 39.94 * 10^-3 meV

(b)

From the above, formula E₀ = 0

Hence, Frequency for transition from J = 0 to J = 1 is (f) = (E₁ - E₀)/ h = E₁ / h = h^'2 / h^' I = h^' / I = h / (4 * pi² * I)

=> f = (6.626 * 10^-34) / (4 * pi² * 1.738 * 10^-45) = 9.657 * 10⁹ Hz = 9.657 GHz

Therefore, frequency of the photon absorbed in the J = 0 to J = 1 transition = 9.657 GHz