The energy of a particle in the second EXCITED state of a harmonic oscillator potential is 5.45 eV.What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s) A)3.31 × 10¹⁵ rad/s B)2.08 × 10¹⁶ rad/s C)4.97 × 10¹⁵ rad/s D)6.95 × 10¹⁵ rad/s

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Quizplus · Accepted Answer

The energy of a particle in the nth excited state of a harmonic oscillator is given by $E_n = (n + \frac{1}{2})\hbar\omega$, where $n$ is the quantum number of the state, $\hbar$ is the reduced Planck's constant, and $\omega$ is the angular frequency. For the second excited state, $n = 2$. Given $E_2 = 5.45 \, \text{eV}$ and converting this energy to joules ($1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$), we have $E_2 = 5.45 \times 1.60 \times 10^{-19} \, \text{J}$. Using $\hbar = 1.055 \times 10^{-34} \, \text{J} \cdot \text{s}$, we can solve for $\omega$.Rearranging the formula to solve for $\omega$ gives $\omega = \frac{2E_n}{(n + \frac{1}{2})\hbar}$. Substituting the given values and solving for $\omega$ yields an angular frequency that matches the value in choice A.

The Energy of a Particle in the Second EXCITED State