Diatomic molecules such as H₂ and O₂ undergo simple harmonic motion with frequencies that obey Hooke's Law, with an effective mass equal to half the atomic mass. If the mass of an oxygen atom is 2.66 × 10⁻²⁶ kg and the observed frequency of oscillation is 4.66 × 10¹³ Hz, what is the effective spring constant associated with the bond between the atoms in an oxygen molecule (O₂)? A) 1140 N/m B) 182 N/m C) 116 N/m D) 29 N/m E) 2280 N/m

Question

Quizplus · Accepted Answer

The effective spring constant (k) can be calculated using the formula for the frequency of oscillation in a simple harmonic oscillator, $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$, where $m$ is the effective mass and $f$ is the frequency of oscillation. Given that the effective mass is half the atomic mass of oxygen, $m = \frac{2.66 \times 10^{-26}}{2} = 1.33 \times 10^{-26}$ kg, and the frequency of oscillation is $4.66 \times 10^{13}$ Hz, we can rearrange the formula to solve for $k$: $k = (2\pi f)^2 \times m$. Substituting the given values, we find $k \approx 1140$ N/m.

Diatomic Molecules Such as H2 and O2 Undergo Simple Harmonic

Diatomic Molecules Such as H₂ and O₂ Undergo Simple Harmonic