Diatomic molecules such as H₂ and O₂ undergo simple harmonic motion with frequencies that obey Hooke's Law with the effective mass being half the atomic mass.If the mass of a hydrogen atom is 1.67 $\times$ 10^-27 kg and the observed frequency of oscillation is 125 $\times$ 10¹² Hz,what is the effective spring constant associated with the bond between the hydrogen atoms? A) 515 N/m B) 164N/m C) 82N/m D) 13N/m E) 1030 N/m

Question

Quizplus · Accepted Answer

The formula for the frequency of simple harmonic motion is:

f = 1/2π * √(k/m)

where k is the spring constant and m is the effective mass.

Since the effective mass is half of the atomic mass (1/2 * 1.67 x 10^-27 kg = 8.35 x 10^-28 kg), we can plug in the given values and solve for k:

125 x 10^12 Hz = 1/2π * √(k/(8.35 x 10^-28 kg))
k = (125 x 10^12 Hz)^2 * 8.35 x 10^-28 kg * 4π^2
k ≈ 1030 N/m

Therefore, the answer is E) 1030 N/m.

Diatomic Molecules Such as H2 and O2 Undergo Simple Harmonic ×\times×

Diatomic Molecules Such as H₂ and O₂ Undergo Simple Harmonic $\times$