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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

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)?

Find the ground state energy (in units of eV) of an electron in a one- dimensional quantum box, if the box is of length \(L=0.100 \mathrm{nm}\).

Electrons from a scanning tunneling microscope encounter a potential barrier that has a height of \(U=4.0 \mathrm{eV}\) above their total energy. By what factor does the tunneling current change if the tip moves a net distance of \(0.10 \mathrm{nm}\) farther from the surface?

Is it possible for the expectation value of the position of an electron to occur at a position where the electron's probability function, \(\Pi(x)\), is zero? If it is possible, give a specific example.

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.
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