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 a hydrogen atom is 1.67 × 10⁻²⁷ kg and the observed frequency of oscillation is 1.25 × 10¹⁴ Hz, what is the effective spring constant associated with the bond between the hydrogen atoms? A) 1030 N/m B) 515 N/m C) 164 N/m D) 82 N/m E) 13 N/m

Question

Quizplus · Accepted Answer

The formula for the frequency of a spring-mass system is:
f = (1/2π) * sqrt(k/m)
where f is the frequency, k is the spring constant, and m is the effective mass.

We know that m = 1/2 * (2 * 1.67 × 10^-27 kg) = 1.67 × 10^-27 kg

Substituting the given frequency in the above formula, we get:
1.25 × 10^14 Hz = (1/2π) * sqrt(k/1.67 × 10^-27 kg)

Simplifying this equation, we can solve for k:
k = (4π^2 * m * f^2)
k = (4π^2 * 1.67 × 10^-27 kg * (1.25 × 10^14 Hz)^2)

k = 515 N/m

Therefore, the effective spring constant associated with the bond between the hydrogen atoms is 515 N/m. The answer is B.

Diatomic Molecules Such as H2 and O2 Undergo Simple Harmonic

Diatomic Molecules Such as H₂ and O₂ Undergo Simple Harmonic