Question

Find the ground state energy (in eV) of a harmonic oscillator with \(\omega=10^{15}\ \text{rad/s}\)
• \(0.5\ \text{eV}\)
• \(0.33\ \text{eV}\)
• \(0.25\ \text{eV}\)
• \(0.66\ \text{eV}\)

• A. \(0.5\ \text{eV}\)
• B. \(0.33\ \text{eV}\)
• C. \(0.25\ \text{eV}\)
• D. \(0.66\ \text{eV}\)

Hint:

Quantum harmonic oscillator: \[ E_n=\left(n+\frac12\right)\hbar\omega \] Ground state: \[ E_0=\frac12\hbar\omega \]

Answer 1

The Correct Option is B

Concept: For a quantum harmonic oscillator, the allowed energy levels are: :contentReference[oaicite:3]{index=3} where:
• \(n=0,1,2,\dots\)
• \(\omega\) = angular frequency
• \(\hbar=\dfrac{h}{2\pi}\) Ground state corresponds to: \[ n=0 \]

Step 1: Write the ground state energy formula. For ground state: \[ n=0 \] Thus: :contentReference[oaicite:4]{index=4}

Step 2: Substitute known values. Given: \[ \omega=10^{15}\ \text{rad/s} \] and: \[ \hbar=1.054\times10^{-34}\ \text{J\,s} \] Therefore: \[ E_0=\frac12(1.054\times10^{-34})(10^{15}) \] \[ E_0=0.527\times10^{-19}\ \text{J} \] \[ E_0=5.27\times10^{-20}\ \text{J} \]

Step 3: Convert joule into electron volt. We know: \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J} \] Thus: \[ E_0=\frac{5.27\times10^{-20}}{1.6\times10^{-19}} \] \[ E_0\approx0.329\ \text{eV} \] \[ \boxed{ E_0\approx0.33\ \text{eV} } \]

Step 4: Choose the correct option. Thus the correct answer is: \[ \boxed{ 0.33\ \text{eV} } \] Hence the correct option is: \[ \boxed{(2)} \] Additional Understanding: Unlike classical oscillators:
• Quantum harmonic oscillators can never have zero energy. Even at ground state: \[ E_0=\frac12\hbar\omega \] This minimum energy is called: \[ \boxed{ \text{Zero point energy} } \] Final Conclusion: Ground state energy is: \[ \boxed{ 0.33\ \text{eV} } \] Hence, the correct answer is: \[ \boxed{(2)} \]