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Question

(C) The average energy $E$ of a one-dimensional harmonic oscillator in thermal equilibrium at temperature $T$ is given by the equipartition theorem as

Vedclass · Accepted Answer

$4.14 	imes 10^{-21} \ J$

Vedclass · Answer

(C) The average energy $E$ of a one-dimensional harmonic oscillator in thermal equilibrium at temperature $T$ is given by the equipartition theorem as $E = k_B T$.However, for a quantum harmonic oscillator, the average energy is given by $E = \frac{h
u}{e^{h
u/k_B T} - 1} + \frac{1}{2}h
u$.In the classical limit where $k_B T \gg h
u$, the average energy is $E = k_B T$.Assuming the question refers to the classical average energy per degree of freedom for a simple harmonic oscillator (which involves both kinetic and potential energy), the total average energy is $E = k_B T$.Given $k_B = 1.38 	imes 10^{-23} \ J K^{-1}$ and $T = 300 \ K$:$E = (1.38 	imes 10^{-23}) 	imes 300 = 4.14 	imes 10^{-21} \ J$.

The average energy possessed by an oscillator at a temperature $300 \ K$ is (Boltzmann constant $= 1.38 \times 10^{-23} \ J K^{-1}$).

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