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 of the oxygen atom is given by:

E = 1/2 kA^2

where k is the effective spring constant and A is the amplitude of oscillation.

Rearranging the equation, we get:

A = sqrt(2E / k)

Substituting the values given in the problem, we get:

A = sqrt((2 x 6.4 x 10^-21 J) / 50 N/m)
  = 1.6 x 10^-11 m

Since each oxygen atom oscillates around the center of mass, the amplitude of oscillation for each oxygen atom is half of this value:

Amplitude of oscillation for each oxygen atom = 0.8 x 10^-11 m = 1.6 x 10^-11 / 2 = 8.0 x 10^-12 m

Therefore, the best choice is B: 1.6 x 10^-11 m.

The Oxygen Molecule (O2)may Be Regarded as Two Masses Connected

The Oxygen Molecule (O₂)may Be Regarded as Two Masses Connected