Question

The oxygen molecule (O2) may be regarded as two masses connected by a spring. In vibrational motion, each oxygen atom alternately approaches, then moves away from the center of mass of the system. If each oxygen atom of mass m = 2.67 ´ 10-26 kg has a vibrational energy of 1.6 ´ 10-21 J and the effective spring constant is 50 N/m, then what is the amplitude of oscillation of each oxygen atom?

Answer 1

2.5 × 10^-11 m

Step-by-step explanation:

The amplitude of oscillation for a simple harmonic oscillator is given by the equation:

A = √(2E / k), where E is the vibrational energy of each oxygen atom and k is the effective spring constant.

Substituting the given values:

A = √(2 * 1.6 × 10^-21 J / 50 N/m)

A = √(3.2 × 10^-21 J / 50 N/m)

A = √(6.4 × 10^-23 m^2 kg / s^2)

A = 2.5 × 10^-11 m