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Chapter 37: Problem 40

An oxygen molecule has a vibrational mode that behaves approximately like a simple harmonic oscillator with frequency \(2.99 \cdot 10^{14} \mathrm{rad} / \mathrm{s} .\) Calculate the energy of the ground state and the first two excited states.

Short Answer

Expert verified
Question: Calculate the energy of an oxygen molecule in its ground state and the first two excited states using a frequency value of \(2.99 \times 10^{14}\,\mathrm{rad/s}\). Answer: The energy of the oxygen molecule in its ground state is \(E_0 = 9.922 \times 10^{-20}\,\mathrm{J}\), in the first excited state is \(E_1 = 29.767 \times 10^{-20}\,\mathrm{J}\), and in the second excited state is \(E_2 = 49.611 \times 10^{-20}\,\mathrm{J}\).

Step by step solution

01

Identify Given Information

We are given the frequency of the oxygen molecule: \(f = 2.99 \times 10^{14}\,\mathrm{rad/s}\). Step 2: Recall the Energy Formula for a Harmonic Oscillator
02

Recall the Energy Formula for a Harmonic Oscillator

The energy of a simple harmonic oscillator can be calculated using the formula: \(E_n = (n+\frac{1}{2})hf\) Where - \(E_n\) is the energy of the \(n\)-th state, - \(n\) is the excited state number (0 for ground state, 1 for the first excited state, and so on), - \(h\) is the Planck constant (\(6.626 \times 10^{-34}\,\mathrm{J\,s}\)), - \(f\) is the frequency of the oscillator. Step 3: Calculate the Energy of the Ground State
03

Calculate the Energy of the Ground State

First, let's calculate the energy of the ground state, which corresponds to \(n=0\): \(E_0 = (\frac{1}{2})hf = \frac{1}{2} \times 6.626 \times 10^{-34}\,\mathrm{J\,s} \times 2.99 \times 10^{14}\,\mathrm{rad/s}\) \(E_0 = 9.922 \times 10^{-20}\,\mathrm{J}\) Step 4: Calculate the Energy of the First Excited State
04

Calculate the Energy of the First Excited State

Next, let's calculate the energy of the first excited state, which corresponds to \(n=1\): \(E_1 = (\frac{3}{2})hf = \frac{3}{2} \times 6.626 \times 10^{-34}\,\mathrm{J\,s} \times 2.99 \times 10^{14}\,\mathrm{rad/s}\) \(E_1 = 29.767 \times 10^{-20}\,\mathrm{J}\) Step 5: Calculate the Energy of the Second Excited State
05

Calculate the Energy of the Second Excited State

Finally, let's calculate the energy of the second excited state, which corresponds to \(n=2\): \(E_2 = (\frac{5}{2})hf = \frac{5}{2} \times 6.626 \times 10^{-34}\,\mathrm{J\,s} \times 2.99 \times 10^{14}\,\mathrm{rad/s}\) \(E_2 = 49.611 \times 10^{-20}\,\mathrm{J}\) Step 6: Present the Results
06

Present the Results

The energy of the oxygen molecule in its ground state is \(E_0 = 9.922 \times 10^{-20}\,\mathrm{J}\), in the first excited state is \(E_1 = 29.767 \times 10^{-20}\,\mathrm{J}\), and in the second excited state is \(E_2 = 49.611 \times 10^{-20}\,\mathrm{J}\).

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Most popular questions from this chapter

Think about what happens to infinite square well wave functions as the quantum number \(n\) approaches infinity. Does the probability distribution in that limit obey the correspondence principle? Explain.

An electron is confined in a three-dimensional cubic space of \(L^{3}\) with infinite potentials. a) Write down the normalized solution of the wave function in the ground state. b) How many energy states are available up to the second excited state from the ground state? (Take the electron spin into account.)

Consider an attractive square-well potential, \(U(x)=0\) for \(x<-\alpha, U(x)=-U_{0}\) for \(-\alpha \leq x \leq \alpha\) where \(U_{0}\) is a positive constant, and \(U(x)=0\) for \(x>\alpha .\) For \(E>0,\) the solution of the Schrödinger equation in the 3 regions will be the following: For \(x<-\alpha, \psi(x)=e^{i \kappa x}+R e^{-i \kappa x}\) where \(\kappa^{2}=2 m E / \hbar^{2}\) and \(R\) is the amplitude of a reflected wave. For \(-\alpha \leq x \leq \alpha, \psi(x)=A e^{i \kappa^{\prime} x}+B e^{-i \kappa^{\prime} x}\) and \(\left(\kappa^{\prime}\right)^{2}=2 m\left(E+U_{0}\right) / \hbar^{2}\). For \(x>\alpha, \psi(x)=T e^{i \kappa x}\) where \(T\) is the amplitude of the transmitted wave. Match \(\psi(x)\) and \(d \psi(x) / d x\) at \(-\alpha\) and \(\alpha\) and find an expression for \(R\). What is the condition for which \(R=0\) (that is, there is no reflected wave)?

An electron is in an infinite square well of width \(a\) : \((U(x)=\infty\) for \(x<0\) and \(x>a\) ). If the electron is in the first excited state, \(\Psi(x)=A \sin (2 \pi x / a),\) at what position is the probability function a maximum? a) 0 b) \(a / 4\) c) \(a / 2\) d) \(3 a / 4\) and \(3 a / 4\) e) at both \(a / 4\)

Determine the lowest 3 energies of the wave function of a proton in a box of width \(1.0 \cdot 10^{-10} \mathrm{~m}\).
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