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 6.4 * 10^-²¹ J and the effective spring constant is 50 N/m, then what is the amplitude of oscillation of each oxygen atom? A) 3.2 *10^-¹¹ m B) 1.6 *10^-¹¹ m C) 1.1 * 10^-¹¹ m D) 8.0 *10^-¹² m

Question

Quizplus · Accepted Answer

The vibrational energy is given by E = (1/2)kA^2, where k is the spring constant and A is the amplitude. Solving for A gives A = sqrt(2E/k). Substituting the given values, A = sqrt(2 * 6.4 x 10^-21 J / 50 N/m) = 1.6 x 10^-11 m.

The Oxygen Molecule (O2) May Be Regarded as Two Masses

The Oxygen Molecule (O₂) May Be Regarded as Two Masses