Question

1. Calculate the “characteristic rotational energy,” ℏ2/2I, for the O2 molecule whose bond length is 0.121 nm.

2. What are the energy and wavelength of photons emitted in an l = 3 to l = 2 transition?

Answers:

1) 1.79×10−4 eV

2) 1.16 mm

Answer 1

(1)

The O₂ molecule rotates about its center of mass. The mass of the two oxygen molecules is the same, the center of mass of the molecules lies halfway between the two oxygen atom. So, each atom is at distance 0.0605nm away from center of mass which is half of 0.121nm (the bond length). The moment of inertia of a point particle of mass m at a distance r away from the axis of rotation is

There are two oxygen atoms, each of which is at distance r=0.0605nm away from the center of mass. The moment of inertia of O₂ is

The characteristic rotational energy is

Substituting

Mass of single Oxygen atom

We get

To convert it to eV we divide it by electronic charge e

(2)

The rotational energy in state l is

The energy of the photon when in l=3 to l=2 transition

The wavelength of the photon is

using 10^-3m=1mm