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Homework 5 Solutions

  • Page ID
    109911
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    Q1

    Calculate the standard deviation of the bond length \(\sigma_X\) of the diatomic molecule \(\ce{^1H^{35}Cl }\) when it is in the ground state and first excited state using the quantum harmonic oscillator wavefunctions. The fundamental harmonic vibrational frequency of \(\ce{HCl }\) is 2989.6 \(cm^{-1}\) and the equilibrium bond length is 0.127 nm. How do you interpret the change in the ratio of average bond length to \(\sigma_X\) as a function of energy in the vibration?

    \(\sigma_X = (\langle x \rangle^{2} - \langle x^{2} \rangle)^{\frac{1}{2}}\)

    \[\psi_0 = (\frac{\mu\omega}{\pi\hbar})^{\frac{1}{4}}*e^{\frac{\mu\omega x^{2}}{2\hbar}}\]

    \[\psi_1 = (\frac{\mu\omega}{\pi\hbar})^{\frac{1}{4}}*\frac{2\mu\omega}{\pi\hbar}*e^{\frac{\mu\omega x^{2}}{2\hbar}}\]

    \(\langle x \rangle = 0\) This is true for both the excited and ground state, as it is always an integral of an even times an odd function over symmetric boundaries, so the standard deviation ends up being merely the square root of \(\langle x^{2} \rangle\).

    For the ground state

    \(\langle x^{2} \rangle = 2(\frac{\mu\omega}{\pi\hbar})^{\frac{1}{2}}\int_0^\infty x^{2}\exp^{\frac{\mu\omega x^{2}}{\hbar}} = (\frac{1}{2}\frac{\hbar}{\mu\omega})^{\frac{1}{2}}) = .6 x 10^{-22} m^{2}\)

    Where \(\frac{\hbar}{\mu * \omega} = 1.2x10^{-22} m^{2}\)

    So the ground state standard deviation is

    \(\sigma_X = 7.7 x 10^{-12} m\)

    For the first excited state

    \(\langle x^{2} \rangle = (\frac{\mu\omega}{\pi\hbar})^{\frac{1}{4}}\frac{4\mu\omega}{\pi\hbar}\int_0^\infty x^{4}\exp^{\frac{\mu\omega x^{2}}{\hbar}} = \frac{3}{2}\frac{\hbar}{\mu\omega} = \frac{3}{2}*1.2x10^{-22}m^{2} = 1.8x10^{-22} m^{2}\)

    So the excited state standard deviation is

    \(\sigma_X = 1.4 x 10^{-11} m\)

    So as expected the standard deviation gets bigger as we move to higher excited states so the ratio of the equilibrium bond length to the standard deviation will go down as the energy of the vibration increases.

    Q2

    What are two requirements for a molecule to absorb IR radiation (via its vibrations)?

    One requirement is that the dipole moment must change between the ground and excited state. Another requirement is that the energy of the IR photon must match the natural frequency of the molecular vibration, stated another way, the incoming photon’s energy has to be such that delta v either increases or decreases by 1 and the transition moment integral must be non-zero.

    Q3

    Using the relevant transition moment integrals. Demonstrate that the probability of a vibration described by a harmonic oscillator in absorbing IR radiation form the \(v=0\) to the \(v=2\) state is forbidden. Is the \(v=1\) to \(v=0\) transition also forbidden?

    To show that a transition is forbidden the transition moment integral must equal zero:

    \[\psi_0 = \left(\dfrac{\alpha}{\pi}\right)^{1/4}e^{-(\alpha x^2)/2}\ \tag{1}\]

    \[\psi_2 = \left(\dfrac{\alpha}{4\pi}\right)^{1/4}(2\alpha x^2 - 1)e^{-(\alpha x^2)/2}\tag{2}\]

    \[\hat\mu = e_cx\tag{3}\]

    This is \(e\) the charge on an electron not \(e\), the natural number, so I added the subscript.

    \[<\psi_0| \langle \hat\mu \rangle |\psi_2> = e_c(\frac{\alpha^2}{4\pi^2})\int_{-\infty}^{\infty}(2\alpha x^3 - x)e^{-\alpha x^2} dx\tag{4}\]

    This integral can be split up to two integrals \(\int_{-\infty}^{\infty}x^3e^{-\alpha x^2}dx\) and \(\int_{-\infty}^{\infty}xe^{-\alpha x^2}dx\) both of these integrals can be seen to be zero as they are odd functions integrated over symmetric limits. Thus the transition moment integral is zero and the transition is not possible.

    \[\psi_1 = \left(\dfrac{4\alpha^3}{\pi}\right)^{1/4}xe^{-(\alpha x^2)/2}\tag{5}\]

    \[<\psi_0|<\hat\mu>|\psi_1> = e_c\int_{-\infty}^{\infty}(\frac{\alpha}{\pi})^{1/4}(\frac{4\alpha^3}{\pi})^{1/4}x^2e^{-\alpha x^2}dx = e(\frac{4\alpha^4}{\pi^2})^{1/2}(\frac{1}{2}(\frac{\pi}{\alpha^3})^{1/2}) = e_c(\frac{1}{2\alpha})^{1/2}\tag{6}\]

    Since this integral is not zero the transition is allowed.

    Q4

    Which of the following molecule absorb in the IR?

    1. \(I_2\) No
    2. \(O_2\) No
    3. \(O_3\) Yes
    4. \(HBr\) Yes
    5. \(HF\) Yes
    6. \(H_2 O\) Yes
    7. \(CD_2\) Yes
    8. \(CO_2\) Yes
    9. \(CH_4\) Yes

    Q5

    What do the presence of overtones in IR spectra reveal about the anharmonicity of the vibration?

    The appearance of overtones reveal that the vibration has anaharmonic character as overtones are not allowed in the harmonic approximation.

    Q6

    What is the energy in cm-1 of a photon of 500 nm energy that may be observed in electron (UV-VIS) spectroscopy? What is the energy of a 6-micron photon typical in IR spectroscopy? What is the energy of a photon absorbed in a typical CO rotation microwave line (\(6 \times 10^{11} Hz\))?

    For each of these problems we need to use

    \[E = h\nu \tag 1 \]

    in the appropriate units.

    \[c = \lambda\nu \tag 2\]

    \[\nu = \dfrac{c}{\lambda} \tag 3 \]

    UV- Vis, 500 nm

    \[E = \dfrac{hc}{\lambda} = \dfrac{1.986\times 10^{-25} \;Jm}{500 \times 10^{-9} \;m} = 3.972 \times 10^{-19} \;J = 19995 \;cm^{-1} \tag 4\]

    IR, 6 micron

    \[E = \dfrac{hc}{\lambda} = \dfrac{1.986\times 10^{-25} \;Jm}{6 \times 10^{-6} \;m} = 3.31 \times 10^{-20} \;J = 1666 \;cm^{-1} \tag 5\]

    Microwave, \(6 \times 10^{11} \;Hz\)

    \[E = h\nu = (6.626 \times 10^{-34} \;Js)(6 \times 10^{11}) \;s^{-1} = 3.97 \times 10^{-22} \;J = 20.0\;cm^{-1} \tag 5\]

    Q7

    Fill in this table.

    Spectroscopic Signature Degree of Freedom
    Type EM Range Typical Wavelength of Transition Typical Energy of Transition sensitive to electronic transition (yes/no) Sensitive to vibrational transition (yes/no) sensitive to rotational transitions (yes/no)
    UV-Visible 500 nm \[4.0 \times 10^-19 \;J\] \[\approx 20,000 \; cm^{-1}\] Y Y, Vibrational fine structure is present in gas phase UV-Vis N
    Infrared 5 \mu m \[3.0 \times 10^-20 \;J\] \[\approx 2,000 \; cm^{-1}\] N Y Y, Rotational fine structure is present in gas phase IR
    Microwave 50 cm

    \[4.0 \times 10^-25 \;J\]

    \[\approx 0.02 \; cm^{-1}\]

    N N Y

    If any spectroscopy is sensitive to more than one degree of freedom, explain why.

    Q8

    The moment of inertia of \(\ce{^1H^{35}Cl }\) is \(2.6 \times 10^{-47} \;Kg\times m^2\). What is the energy for rotation for \(\ce{^1H^{35}Cl }\) in the \(J=5\) and \(J=20\) states? For a molecule to be thermally excited, the energy of the eigenstate must be comparable to \(k_bT\), with \(k_b\) as the Boltzmann's constant and \(T\) is absolute temperature. What temperature is needed for the \(J=5\) and \(J=20\) rotational states of \(\ce{^1H^{35}Cl }\) to be thermally occupied? (Hint: assuming the term "comparable" is "equal" for this problem).

    Energy of a rigid rotor is

    \[E = BJ(J+1) \;where B = \dfrac{\hbar^2}{2I} \tag 1\]

    for \(J = 5 \;and J = 20\)

    \[E =\dfrac{J\hbar^2}{2I}(J+1) = \dfrac{5\hbar^2}{2I}(5+1) =\dfrac{30 \times (1.054 \times 10^{-34} \; Js)^2 }{2 \times 2.6 \times 10^{-47} \;Kg\times m^2} = 6.409 \times 10^{-20} J = 322 cm^{-1} \tag 2\]

    \[E =\dfrac{J\hbar^2}{2I}(J+1) = \dfrac{20\hbar^2}{2I}(20+1) =\dfrac{420 \times (1.054 \times 10^{-34} \; Js)^2 }{2 \times 2.6 \times 10^{-47} \;Kg\times m^2} = 8.972 \times 10^{-20} J = 4516 cm^{-1} \tag 3\]

    For the temperature

    \[E=k_bT = E_{J=5} \tag4\]

    \[T = \dfrac{E_{J=5}}{k_b} = \dfrac{322 cm^{-1}}{0.69 cm^{-1} \times K^{-1}} = 466 K \tag{5}\]

    \[T = \dfrac{E_{J=20}}{k_b} = \dfrac{4516 cm^{-1}}{0.69 cm^{-1} \times K^{-1}} = 6544 K \tag{6}\]

    Q9

    \(\ce{^1H^{35}Cl }\) has a bond length of 0.12746 nm and fundamental stretching vibration at 2,886 cm-1. What is the temperature required for the \(v=1\) mode to be thermally excited? (Hint: assuming the term "comparable" is "equal" for this problem).

    The bond length is extra information you don't need. This problem is a plug and chug just like the last part of #8.

    \[T = \dfrac{E_{v}}{k_b} = \dfrac{2886 cm^{-1}}{0.69 cm^{-1} \times K^{-1}} = 4182 K \tag{1}\]

    Q10

    \(\ce{^1H^{19}F}\) has an equilibrium bond length of 91.7 pm and a spring constant of 970 N/m. The molecule rotates freely in a three-dimensional space as a gas.

    1. What is the zero point energy associated with this rotation? Will this differ if you were considering only vibration?
    2. What is the lowest energy microwave transition observed absorbed \(\ce{^1H^{19}F}\) ascribed to rotational motion (assuming a rigid rotor described the rotation)?

      a) Rotations are allowed to have J=0, so the zero point energy = 0. Vibrations have zero point energy and the value would be determined by the spring constant.

      \[\omega = \sqrt{\dfrac{k}{\mu}} \tag{1}\]

      \[\omega = \sqrt{\dfrac{970 \;N\times m^{-1}}{1.577 \times 10^{-27} \;kg}} = 7.84 \times 10^{14} \; Hz \tag{2}\]

      \[E = (v +1/2)\hbar\omega = \dfrac{\hbar\omega}{2} = \dfrac{1.054 \times 10^{-34} \;Js \times 7.84 \times 10^{14} \; Hz}{2} = 4.13 \times 10^{-20} = 2080 \;cm^{-1} \tag{3}\]

      b) The rotational transition will be from J=0 to J=1

      \[E = BJ(J+1) \;Where B = \dfrac{\hbar^2}{2I} \;Where I = r^2\mu \tag {4}\]

      \[E = \dfrac{2\hbar^2}{2r^2\mu} = \dfrac{2 \times (1.054 \times 10^{-34} \; Js) ^2}{2 \times (91.7 \times 10^{-12} \; m)^2 \times 1.577 \times 10^{-27} \; kg} = 8.377 \times 10^{-22} = 42 \;cm^{-1} \tag{5}\]


    Homework 5 Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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