Question

Consider the diatomic molecule oxygen O2 which is rotating in the xy plane about the z axis passing through its center, perpendicular to its length. The mass of each oxygen atom is 2.66 × 10−26 kg, and at room temperature, the average separation distance between the two oxygen atoms is 2.2 × 10−10 m (treat the atoms as point masses). If the angular speed of the molecule about the z axis is 5.49 × 1012 rad/s, what is its rotational kinetic energy? Answer in units of J.

Answer 1

Answer: 9.7*10^-21 J

Step-by-step explanation:

Given

Mass of oxygen atom, m = 2.66*10^-26 kg

Separation distance between the two atoms, r = 2.2*10^-10 m

Angular speed, w = 5.49*10^12 rad/s

First, we start by finding our moment of Inertia, I, using the relation

I = 1/2mr²

I = 1/2 * 2.66*10^-26 * (2.2*10^-10)²

I = 1/2 * 1.2874*10^-45

I = 6.437*10^-46 kgm²

The Rotational Kinetic Energy is given by

KE = 1/2Iw²

Now, using this value of I, we plug it in the energy equation.

KE = 1/2 * 6.437*10^-46 * (5.49*10^12)²

KE = 1/2 * 1.94*10^-20

KE = 9.7*10^-21 J

Thus, the rotational kinetic energy is 9.7*10^-21 J