 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/a8/solution

E_{J} = ℏ^{2} J (J + 1) / (2 I)

and electric-dipole absorption has

Δ J = + 1

. A line labelled

J^{'}

therefore has

ν_{J^{'}} = \frac{h J ^{'}}{4 π ^{2} I}, Δ ν = \frac{h}{4 π ^{2} I} .

The original graph gives adjacent spacing about

0.115 THz

; for example the

J^{'} = 5

line is near

0.575 THz

. Thus

I ≃ 1.46 \times 1 0^{- 46} kg m^{2}

. Using the reduced mass,

μ = \frac{12 \times 16}{12 + 16} m_{H} = 1.145 \times 1 0^{- 26} kg, r = I / μ ≃ 1.13 \times 1 0^{- 10} m = 0.113 nm .

Centrifugal distortion of a nonrigid molecule explains small departures from equal spacing at higher

J

 New to topics? Read the docs here!