Diatomic molecules such as $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ 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 $1.67 \times 10^{-27} \mathrm{~kg}$ and the observed frequency of oscillation is $1.25 \times 10^{14} \mathrm{~Hz}$, what is the effective spring constant associated with the bond between the hydrogen atoms?
A) $515 \mathrm{~N} / \mathrm{m}$
B) $13 \mathrm{~N} / \mathrm{m}$
C) $164 \mathrm{~N} / \mathrm{m}$
D) $82 \mathrm{~N} / \mathrm{m}$
E) $1030 \mathrm{~N} / \mathrm{m}$

Question

Quizplus · Accepted Answer

The effective spring constant $k$ can be calculated using the formula for the frequency of a simple harmonic oscillator, $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$, where $m$ is the effective mass of the diatomic molecule, which is half the mass of one atom for diatomic molecules like $H_2$. Given the mass of a hydrogen atom is $1.67 \times 10^{-27} \mathrm{kg}$, the effective mass $m = \frac{1}{2} \times 1.67 \times 10^{-27} \mathrm{kg} = 0.835 \times 10^{-27} \mathrm{kg}$. Rearranging the formula to solve for $k$ and substituting the given frequency $f = 1.25 \times 10^{14} \mathrm{Hz}$, we find $k = (2\pi f)^2 \times m$, which calculates to approximately $515 \mathrm{N/m}$.

Diatomic Molecules Such As H2\mathrm{H}_{2}H2​ And O2\mathrm{O}_{2}O2​ Undergo Simple Harmonic Motion with Frequencies That Obey Hooke's Law

Diatomic Molecules Such As $\mathrm{H}_{2}$ And $\mathrm{O}_{2}$ Undergo Simple Harmonic Motion with Frequencies That Obey Hooke's Law