# rotational energy levels

Posted on Ene 1, 2021

The spherical harmonics called $$Y_J^{m_J}$$ are functions whose probability $$|Y_J^{m_J}|^2$$ has the well known shapes of the s, p and d orbitals etc learned in general chemistry. So, although the internuclear axis is not always aligned with the z-axis, the probability is highest for this alignment. There is only, $$J=1$$: The next energy level is $$J = 1$$ with energy $$\dfrac {2\hbar ^2}{2I}$$. The rotation transition refers to the loss or gain … For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Describe how the spacing between levels varies with increasing $$J$$. However, for many actual diatomics this model is too restrictive since distances are usually not completely fixed and corrections on the rigid model can be made to compensate for small variations in the distance. Atkins, Peter and de Paula, Julio. This lecture is in continuation of our series on Rotational Spectroscopy. \frac{d^{2}}{d \varphi^{2}} \Phi_{\mathrm{m}}(\varphi)+m_{J}^{2} \Phi_{\mathrm{m}}(\varphi)=& \frac{d}{d \varphi}\left(\mathrm{N}\left(\pm \mathrm{i} m_{J}\right) e^{\pm \mathrm{i} m_{J} \varphi}\right)+m_{J}^{2} \Phi_{\mathrm{m}}(\varphi) \\ For a diatomic molecule the energy difference between rotational levels (J to J+1) is given by: EJ + 1 − EJ = B(J + 1)(J + 2) − BJ(J = 1) = 2B(J + 1) with J=0, 1, 2,... Because the difference of energy between rotational levels is in the microwave region (1-10 cm -1) rotational spectroscopy is commonly called microwave spectroscopy. Rotational spectroscopy. Finding the $$\Theta (\theta)$$ functions that are solutions to the $$\theta$$-equation (Equation $$\ref{5.8.18}$$) is a more complicated process. Note this diagram is not to scale. In this discussion we’ll concentrate mostly on diatomic molecules, to keep things as simple as possible. We first write the rigid rotor wavefunctions as the product of a theta-function depending only on $$\theta$$ and a phi-function depending only on $$\varphi$$, $| \psi (\theta , \varphi ) \rangle = | \Theta (\theta ) \Phi (\varphi) \rangle \label {5.8.11}$, We then substitute the product wavefunction and the Hamiltonian written in spherical coordinates into the Schrödinger Equation $$\ref{5.8.12}$$, $\hat {H} | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta ) \Phi (\varphi) \rangle \label {5.8.12}$, $-\dfrac {\hbar ^2}{2\mu r^2_0} \left [ \dfrac {\partial}{\partial r_0} r^2_0 \dfrac {\partial}{\partial r_0} + \dfrac {1}{\sin \theta} \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta) \Phi (\varphi) \rangle \label {5.8.13}$, Since $$r = r_0$$ is constant for the rigid rotor and does not appear as a variable in the functions, the partial derivatives with respect to $$r$$ are zero; i.e. This rotating molecule can be assumed to be a rigid rotor molecule. Often $$m_J$$ is referred to as just $$m$$ for convenience. That is, from J = 0 to J = 1, the ΔE0 → 1 is 2Bh and from J = 1 to J = 2, the ΔE1 → 2 is 4Bh. The rotational spectra of non-polar molecules cannot be observed by those methods, but can be observed and measured by Raman spectroscopy. [ "article:topic", "rigid rotor", "cyclic boundary condition", "spherical harmonics", "showtoc:no", "license:ccbyncsa" ], 5.7: Hermite Polynomials are either Even or Odd Functions, 5.9: The Rigid Rotator is a Model for a Rotating Diatomic Molecule, Copenhagen interpretation of wavefunctions, information contact us at info@libretexts.org, status page at https://status.libretexts.org, $$\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$$, $$\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{i \varphi}$$, $$\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$$, $$\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{-i \varphi}$$, $$\sqrt {\dfrac {5}{8}}(3\cos ^2 \theta - 1)$$, $$\sqrt {\dfrac {5}{16\pi}}(3\cos ^2 \theta - 1)$$, $$\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$$, $$\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{i\varphi}$$, $$\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{-i\varphi}$$, $$\sqrt {\dfrac {15}{16}} \sin ^2 \theta$$, $$\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$$, $$\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{2i\varphi}$$, $$\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{-2i\varphi}$$, Compare the classical and quantum rigid rotor in three dimensions, Demonstrate how to use the Separation of Variable technique to solve the 3D rigid rotor Schrödinger Equation, Identify and interpret the two quantum numbers for a 3D quantum rigid rotor including the range of allowed values, Describe the wavefunctions of the 3D quantum rigid rotor in terms of nodes, average displacements and most probable displacements, Describe the energies of the 3D quantum rigid rotor in terms of values and degeneracies, $$J=0$$: The lowest energy state has $$J = 0$$ and $$m_J = 0$$. This conclusion means that molecules are not rotating in the classical sense, but they still have some, but not all, of the properties associated with classical rotation. Hello members, I have a doubt. $$J$$ can be 0 or any positive integer greater than or equal to $$m_J$$. Equation $$\ref{5.8.29}$$ means that $$J$$ controls the allowed values of $$m_J$$. The $$\varphi$$-equation is similar to the Schrödinger Equation for the free particle. To solve the Schrödinger equation for the rigid rotor, we will separate the variables and form single-variable equations that can be solved independently. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. H-H and Cl-Cl don't give rotational spectrum (microwave inactive). Molecules can also undergo transitions in their vibrational or rotational energy levels. ROTATIONAL ENERGY LEVELS AND ROTATIONAL SPECTRA OF A DIATOMIC MOLECULE || RIGID ROTATOR MODEL || Pankaj Physics Gulati. Figure 7.5.1: Energy levels and line positions calculated in the rigid rotor approximation. The combination of Equations $$\ref{5.8.16}$$ and $$\ref{5.8.28}$$ reveals that the energy of this system is quantized. Rotational transitions of molecules refer to the abrupt change in the angular momentum of that molecule. Rotational energy levels depend only on the momentum of inertia I and the orbital angular momentum quantum number $$l$$ (in this case, $$l = 0$$, 1, and 2). Rotational spectroscopy is sometimes referred to as pure rotational spectroscop… If a diatomic molecule is assumed to be rigid (i.e., internal vibrations are not considered) and composed of two atoms…. Write a paragraph describing the information about a rotating molecule that is provided in the polar plot of $$Pr [\theta, \theta ]$$ for the $$J = 1$$, $$m_J = \pm 1$$ state in Figure $$\PageIndex{1}$$. Exercise $$\PageIndex{5}$$: Cyclic Boundary Conditions. Carry out the steps leading from Equation $$\ref{5.8.15}$$ to Equation $$\ref{5.8.17}$$. $E = \dfrac {\hbar^2}{I} = \dfrac {\hbar^2}{\mu r^2} \nonumber$, $\mu_{O2} = \dfrac{m_{O} m_{O}}{m_{O} + m_{O}} = \dfrac{(15.9994)(15.9994)}{15.9994 + 15.9994} = 7.9997 \nonumber$. Use Euler’s Formula to show that $$e^{im_J2\pi}$$ equals 1 for $$m_J$$ equal to zero or any positive or negative integer. And the relevant Schrodinger equation that we need to solve in order to get the allowed energy levels is called the rigid-rotator equation. &=-\mathrm{N} m_{J}^{2} e^{\pm i m_{J} \varphi}+\mathrm{N} m_{J}^{2} e^{\pm i m_{J} \varphi}=0 Substitute Equation $$\ref{5.8.22}$$ into Equation $$\ref{5.8.21}$$ to show that it is a solution to that differential equation. In the classical picture, a molecule rotating in a plane perpendicular to the xy‑plane must have the internuclear axis lie in the xy‑plane twice every revolution, but the quantum mechanical description says that the probability of being in the xy-plane is zero. From solving the Schrödinger equation for a rigid rotor we have the relationship for energies of each rotational eigenstate (Equation \ref{5.8.30}): Using this equation, we can plug in the different values of the $$J$$ quantum number so that. The energy is $$\dfrac {6\hbar ^2}{2I}$$, and there are, For J=2, $$E = (2)(3)(ħ^2/2I) = 6(ħ^2/2I)$$, For J=3, $$E = (3)(4)(ħ^2/2I) = 12(ħ^2/2I)$$, For J=4, $$E = (4)(5)(ħ^2/2I) = 20(ħ^2/2I)$$, For J=5, $$E = (5)(6)(ħ^2/2I) = 30(ħ^2/2I)$$. They are oriented so that the products of inertia are zero. Within the Copenhagen interpretation of wavefunctions, the absolute square (or modulus squared) of the rigid rotor wavefunction $$Y^{m_{J*}}_J (\theta, \varphi) Y^{m_J}_J (\theta, \varphi)$$ gives the probability density for finding the internuclear axis oriented at $$\theta$$ to the z-axis and $$\varphi$$ to the x-axis. Since $$V=0$$ then $$E_{tot} = T$$ and we can also say that: $T = \dfrac{1}{2}\sum{m_{i}v_{i}^2} \label{5.8.3}$. The two-dimensional space for a rigid rotor is defined as the surface of a sphere of radius $$r_0$$, as shown in Figure $$\PageIndex{2}$$. For each state with $$J = 0$$ and $$J = 1$$, use the function form of the $$Y$$ spherical harmonics and Figure $$\PageIndex{1}$$ to determine the most probable orientation of the internuclear axis in a diatomic molecule, i.e., the most probable values for $$\theta$$ and $$\theta$$. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We call this constant $$m_J^2$$ because soon we will need the square root of it. A rigid, nonlinear molecule has rotational energy levels determined by three rotational constants, conventionally written ,, and , which can often be determined by rotational spectroscopy. For a fixed value of $$J$$, the different values of $$m_J$$ reflect the different directions the angular momentum vector could be pointing – for large, positive $$m_J$$ the angular momentum is mostly along +z; if $$m_J$$ is zero the angular momentum is orthogonal to $$z$$. …radiation can cause changes in rotational energy levels within molecules, making it useful for other purposes. Keep in mind that, if $$y$$ is not a function of $$x$$, $\dfrac {dy}{dx} = y \dfrac {d}{dx} \nonumber$, Equation $$\ref{5.8.17}$$ says that the function on the left, depending only on the variable $$\theta$$, always equals the function on the right, depending only on the variable $$\varphi$$, for all values of $$\theta$$ and $$\varphi$$. 44-4 we picture a diatomic molecule as a rigid dumbbell (two point masses m, and mz separated by a constant distance r~ that can rotate about axes through its center of mass, perpendicular to the line joining them. kinldy clear it. A wavefunction that is a solution to the rigid rotor Schrödinger Equation (Equation $$\ref{5.8.11}$$) can be written as a single function $$Y(\theta, \varphi)$$, which is called a spherical harmonic function. Well, i calculated the moment of inertia, I=mr^2; m is the mass of the object. The first term in the above nuclear wave function equation corresponds to kinetic energy of nuclei due to their radial motion. The Spherical Harmonic for this case is, $Y^0_1 = \sqrt{ \dfrac {3}{4 \pi}} \cos \theta \label {5.8.34}$. $E = \dfrac {\hbar ^2 \lambda}{2I} = J(J + 1) \dfrac {\hbar ^2}{2I} \label {5.8.30}$. \begin{align} e^{im_J \varphi} &= e^{im_J (\varphi + 2\pi)} \label{5.8.24} \\[4pt] &= e^{im_J\varphi} e^{im_J2\pi} \label {5.8.25} \end{align}, For the equality in Equation $$\ref{5.8.25}$$ to hold, $$e^{i m_J 2 \pi}$$ must equal 1, which is true only when, $m_J = \cdots , -3, -2, -1, 0, 1, 2, 3, \cdots \label {5.8.26}$. Values for $$m$$ are found by using a cyclic boundary condition. Since $$\omega$$ is a scalar constant, we can rewrite Equation \ref{5.8.6} as: $T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2} \label{5.8.7}$. Sketch this region as a shaded area on Figure $$\PageIndex{1}$$. A rigid rotor only approximates a rotating diatomic molecular if vibration is ignored. Term Φ s | N 2 |Φ s / 2μR 2 represents rotational kinetic energy of the two nuclei, about their center of mass, in a given electronic state Φ s. Possible values of the same are different rotational energy levels for the molecule. The $$\Theta (\theta)$$ functions, along with their normalization constants, are shown in the third column of Table $$\PageIndex{1}$$. A caroussel of mass 1 tonn( 1000 kg)(evenly distributed to the disc) has a diameter 20m and rotates 10 times per minute. p. 515. Construct a rotational energy level diagram including $$J = 0$$ through $$J=5$$. Be on the lookout for your Britannica newsletter to get trusted stories delivered right to your inbox. Missed the LibreFest? Effect of anharmonicity. The fixed distance between the two masses and the values of the masses are the only characteristics of the rigid model. Inserting $$\lambda$$, evaluating partial derivatives, and rearranging Equation $$\ref{5.8.15}$$ produces, $\dfrac {1}{\Theta (\theta)} \left [ \sin \theta \dfrac {\partial}{\partial \theta } \left (\sin \theta \dfrac {\partial}{\partial \theta } \right ) \Theta (\theta) + \left ( \lambda \sin ^2 \theta \right ) \Theta (\theta) \right ] = - \dfrac {1}{\Phi (\varphi)} \dfrac {\partial ^2}{\partial \varphi ^2} \Phi (\varphi) \label {5.8.17}$. Solutions are found to be a set of power series called Associated Legendre Functions (Table M2), which are power series of trigonometric functions, i.e., products and powers of sine and cosine functions. The rotational energy levels within a molecule correspond to the different possible ways in which a portion of a molecule can revolve around the chemical bond that binds it to the remainder of the…, In the gas phase, molecules are relatively far apart compared to their size and are free to undergo rotation around their axes. Rotational energy levels of a diatomic molecule Spectra of a diatomic molecule Moments of inertia for polyatomic molecules Polyatomic molecular rotational spectra Intensities of microwave spectra Sample Spectra Problems and quizzes Solutions Topic 2 Rotational energy levels of diatomic molecules A molecule rotating about an axis with an angular velocity C=O (carbon monoxide) is an example. In spherical coordinates the area element used for integrating $$\theta$$ and $$\varphi$$ is, $ds = \sin \theta\, d \theta \,d \varphi \label {5.8.33}$. Rotational–vibrational spectroscopy is a branch of molecular spectroscopy concerned with infrared and Raman spectra of molecules in the gas phase. We need to evaluate Equation \ref{5.8.23} with $$\psi(\varphi)=N e^{\pm i m J \varphi}$$, \begin{align*} \psi^{*}(\varphi) \psi(\varphi) &= N e^{+i m J \varphi} N e^{-i m J \varphi} \\[4pt] &=N^{2} \\[4pt] 1=\int_{0}^{2 \pi} N^* N d \varphi=1 & \\[4pt] N^{2} (2 \pi) =1 \\[4pt] N=\sqrt{1 / 2 \pi} \end{align*}. Energy level transitions can also be nonradiative, meaning emission or absorption of a photon is not involved. The energies of the rotational levels are given by Equation 7.6.5, E = J(J + 1)ℏ2 2I and each energy level has a degeneracy of 2J + 1 due to the different mJ values. The momentum of inertia depends, in turn, on the equilibrium separation distance (which is given) and the reduced mass, which depends on the masses of the H and Cl atoms. apart while the rotational levels have typical separations of 1 - 100 cm-1 The solution to the $$\theta$$-equation requires that $$λ$$ in Equation $$\ref{5.8.17}$$ be given by. We can rewrite Equation $$\ref{5.8.3}$$ as, $T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2 \label{5.8.10}$. Dening the rotational constant as B=~2 2r2 1 hc= h 8ˇ2cr2, the rotational terms are simply F(J) = BJ(J+ 1): In a transition from a rotational level J00(lower level) to J0(higher level), … The range of the integral is only from $$0$$ to $$2π$$ because the angle $$\varphi$$ specifies the position of the internuclear axis relative to the x-axis of the coordinate system and angles greater than $$2π$$ do not specify additional new positions. Benjamin, Inc, pg.91-100. the functions do not change with respect to $$r$$. Energy level representations of the rotation–vibration transitions in a heteronuclear diatomic molecule, shown in order of increasing optical frequency and mapped to the corresponding lines in the absorption spectrum. Diatomics. The energies of the spectral lines are 2(J+1)B for the transitions J -> J+1. Vibrational spectroscopy. We also can substitute the symbol $$I$$ for the moment of inertia, $$\mu r^2_0$$ in the denominator of the left hand side of Equation $$\ref{5.8.13}$$, to give, $-\dfrac {\hbar ^2}{2I} \left [ \dfrac {1}{\sin \theta} \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2}\right ] | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta) \Phi (\varphi) \rangle \label {5.8.14}$, To begin the process of the Separating of Variables technique, multiply each side of Equation $$\ref{5.8.14}$$ by $$\dfrac {2I}{\hbar ^2}$$ and $$\dfrac {-\sin ^2 \theta}{\Theta (\theta) \Phi (\varphi)}$$ to give, $\dfrac {1}{\Theta (\theta) \psi (\varphi)} \left [ \sin \theta \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {\partial ^2}{\partial \varphi ^2}\right ] \Theta (\theta ) \Phi (\varphi) = \dfrac {-2IE \sin ^2 \theta}{\hbar ^2} \label {5.8.15}$. It is convenient to discuss rotation with in the spherical coordinate system rather than the Cartesian system (Figure $$\PageIndex{1}$$). Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy. The rotational quantum numbers in the ground and first excited vibrational levels are here designated J and J', respectively. Note that a double integral will be needed. In Fig. Rotational energy levels. Construct a rotational energy level diagram including $$J = 0$$ through $$J=5$$. For a transition to occur between two rotational energy levels of a diatomic molecule, it must possess a permanent dipole moment (this requires that the two atoms be different), the frequency of the radiation incident on the molecule must satisfy the quantum condition E J ′ − E J = hν, and the selection rule ΔJ = ±1 must be obeyed. The spectra of polar molecules can be measured in absorption or emission by microwave spectroscopy or by far infrared spectroscopy. &\left.=\mathrm{N}\left(\pm \mathrm{i} m_{J}\right)^{2} e^{\pm i m_{J} \varphi}\right)+m_{J}^{2}\left(\mathrm{N} e^{\pm \mathrm{i} m_{J} \varphi}\right) \\ Vibration-rotation spectra. Physically, the energy of the rotation does not depend on the direction, which is reflected in the fact that the energy depends only on $$J$$ (Equation $$\ref{5.8.30}$$), which measures the length of the vector, not its direction given mb $$m_J$$. Anderson, J.M. $\Phi _{m_J} (\varphi) = \sqrt{\dfrac{1}{2\pi}} e^{\pm i m_J \varphi} \nonumber$. By signing up for this email, you are agreeing to news, offers, and information from Encyclopaedia Britannica. \end{aligned}\]. Claculate the rotational energy levels and angular quantum number. The rotational kinetic energy is determined by the three moments-of-inertia in the principal axis system. The relationship between the three moments of inertia, and hence the energy levels, depends … The partial derivatives have been replaced by total derivatives because only a single variable is involved in each equation. Show how Equations $$\ref{5.8.18}$$ and $$\ref{5.8.21}$$ are obtained from Equation $$\ref{5.8.17}$$. New York: W.H. Calculate $$J = 0$$ to $$J = 1$$ rotational transition of the $$\ce{O2}$$ molecule with a bond length of 121 pm. Describe how the spacing between levels varies with increasing $$J$$. David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski ("Quantum States of Atoms and Molecules"). Only two variables $$\theta$$ and $$\varphi$$ are required in the rigid rotor model because the bond length, $$r$$, is taken to be the constant $$r_0$$. Rotational energy levels – polyatomic molecules. Also, since the probability is independent of the angle $$\varphi$$, the internuclear axis can be found in any plane containing the z-axis with equal probability. They have moments of inertia Ix, Iy, Izassociated with each axis, and also corresponding rotational constants A, B and C [A = h/(8 2cIx), B = h/(8 2cIy), C = h/(8 2cIz)]. Energy level diagram of a diatomic molecule showing the n = 0and n = 1 vibrational energy levels and associated rotational states. It is concerned with transitions between rotational energy levels in the molecules, the molecule gives a rotational spectrum only If it has a permanent dipole moment: A‾ B+ B+ A‾ Rotating molecule H-Cl, and C=O give rotational spectrum (microwave active). These functions are tabulated above for $$J = 0$$ through $$J = 2$$ and for $$J = 3$$ in the Spherical Harmonics Table (M4) Polar plots of some of the $$\theta$$-functions are shown in Figure $$\PageIndex{3}$$. Of air and for water calculated in the new year with a Britannica Membership - Now 30 %.. Hence, there exist \ ( m_J\ ) can equal any positive integer greater than or equal to (... 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