Solution
Ground state energy of an Oscillator is
E0=ℏω2=10×10−3eVE_0=\frac{\hbar \omega}{2}=10\times10^{-3}eVE0=2ℏω=10×10−3eV
First excited state is given
E1=3ℏω2=3×10−2eVE_1=\frac{3\hbar \omega}{2}=3\times10^{-2}eVE1=23ℏω=3×10−2eV
Thermal energy is given
E=kT=1.38×10−23×300E= kT=1.38\times10^{-23}\times300E=kT=1.38×10−23×300
E=4.14×10−23eVE=4.14\times10^{-23}eVE=4.14×10−23eV
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