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Chapter 35: Problem 23

Find the ground-state energy for a particle in a harmonic oscillator potential whose classical angular frequency \(\omega\) is \(1.0 \times 10^{17} \mathrm{s}^{-1}.\)

Short Answer

Expert verified
The ground-state energy for a particle in a harmonic oscillator potential with angular frequency of \(1.0 \times 10^{17} \mathrm{s}^{-1}\) is approximately \(5.27 \times 10^{-18}\) joules.

Step by step solution

01

Identify the Knowns

Recognize the known values from the question. This is important to understand which values to substitute in the equation for ground state energy. Here, the angular frequency \( \omega \) is known as \( 1.0 \times 10^{17} s^{-1} \). The quantum number \( n \) will be 0 for ground state energy.
02

Utilize the Ground-State Energy Equation

Rewrite the ground-state energy equation, \(E = \hbar\omega(n+\frac{1}{2}) \), and replace \( n \) with 0 since \( n \) is 0 for ground state energy. This results in the equation \(E = \hbar\omega(\frac{1}{2}) \) .
03

Substitute Known Values

Substitute the known values into the equation. The value of reduced Planck's constant \(\hbar \) is \( 1.0545718 \times 10^{-34} m^2 kg/s \) and the angular frequency \( \omega \) given is \( 1.0 \times 10^{17} s^{-1} \). Hence, the equation becomes \( E = (1.0545718 \times 10^{-34} m^2 kg/s) \times (1.0 \times 10^{17} s^{-1}) \times (\frac{1}{2}) \).
04

Calculate the Ground-State Energy

Finally, calculate the answer by performing the multiplication operation given in the equation. The \( s^{-1} \) in \(\hbar\) and \(\omega\) get cancelled out. This will give the ground state energy.

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Most popular questions from this chapter

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