Suppose the natural frequency of oscillation for H₂ is $\backsim$10¹² Hz,the effective spring constant between the two H atoms is of the order

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The natural frequency of a diatomic molecule like H2 is related to the spring constant k and reduced mass μ by the formula f = (1/2π)√(k/μ). Given f ≈ 10^12 Hz and using the reduced mass of H2, the spring constant k is calculated to be approximately 10^-1 N/m.

Suppose the Natural Frequency of Oscillation for H2 Is ∽\backsim∽ 1012 Hz,the Effective Spring Constant Between the Two H Atoms

Suppose the Natural Frequency of Oscillation for H₂ Is $\backsim$ 10¹² Hz,the Effective Spring Constant Between the Two H Atoms