Tiêu đề Molecular Spectroscopy.pdf ✅

University College of Science, Tumkur 1 MOLECULAR SPECTROSCOPY Apart from the ground state, a molecule can be in a higher energy state, and transitions between the various energy levels give rise to the observed molecular spectra. The energy of a diatomic molecule arises from three modes: (i) the electronic configuration of the electrons in the molecule, (ii) the vibration of the atoms about the equilibrium position, and (iii) the rotation of the molecule as a whole about its centre of mass. The total energy of a molecule can be expressed as the sum of three independent terms: E = Ee + Ev + Er where, Ee is the electronic energy, Ev is the vibrational energy and Er is the rotational energy of the molecule. In the above expression, Ee , Ev and Er are quantized and Ee > Ev > Er . If the molecule remains in its ground state level of electronic and vibrational energies, and if it suffers transitions between different rotational energies, we get the pure rotational spectrum of the molecule. These spectra are in the microwave and far infrared regions. If the molecule remains in its ground state level of electronic energy, and if transitions occur between different vibrational and rotational energies we get the vibration-rotation spectrum of the molecule. These spectra are in the near infrared region. If transitions occur between different electronic-vibrational-rotational energy levels, we get the electronic spectrum of the molecule. These spectra are in the visible and ultraviolet regions. Theory of the Origin of Pure Rotational Spectrum of a Molecule: Consider a diatomic molecule consisting of atoms of masses m1 and m2 separated by a distance r0 (bond length). The diatomic molecule can rotate about its centre of mass C as shown in figure. Let r1 and r2 be the distances of m1 and m2, respectively from the centre of mass. From the figure, r0 = r1 + r2
University College of Science, Tumkur 2 The moment of inertia of this molecule about an axis passing through its centre of mass and perpendicular to a line joining the atoms is I = m1r1 2 + m2r2 2 From the definition of centre of mass, moment of masses of atoms about the centre of mass is zero. i.e. m1r1 − m2r2 = 0 m1r1 = m2r2 (1) Therefore, I = m2r2r1 + m1r1r2 I = r2r1 (m2 + m1 ) (2) We have, r0 = r1 + r2 Or, r2 = r0 − r1 Substituting in equation (1), m1r1 = m2r2 = m2 (r0 − r1 ) Solving for r1 and r2, m1r1 = m2 (r0 − r1 ) m1r1 = m2r0 − m2r1 This implies, m1r1 + m2r1 = m2r0 r1 (m1 + m2 ) = m2r0 Therefore, r1 = m2r0 (m1 + m2 ) Similarly, r2 = m1r0 (m1 + m2 ) Substituting r1 and r2 in equation (2), I = m2r0 (m1 + m2 ) × m1r0 (m1 + m2 ) × (m2 + m1 ) I = m1m2r0 2 (m1 + m2 ) Or, I = μr0 2 (3)
University College of Science, Tumkur 3 Where, μ = m1m2 (m1+m2 ) is called reduced mass of the system. Equation (3) gives the moment of inertia of diatomic molecule in terms of atomic masses and bond length. The angular momentum of the diatomic molecule is related to moment of inertia as, L = Iω Where, ω is angular velocity of rotating molecule. The angular momentum is quantized and represented as, L = √J(J + 1) h 2π Where, h is Planck’s constant and J Is called rotational quantum number. J takes integer values, J = 0, 1, 2, 3 ... ... Now, the rotational kinetic energy is given by, E = 1 2 Iω 2 Expressing interms of L, E = L 2 2I i.e. E = J(J + 1)h 2 8π 2I joules The above expression gives the allowed quantized rotational energy values in joules. In the study of spectra, its is required to consider the difference between energies of energy levels, particularly, corresponding frequency, ν = ΔE h in Hz or wavenumber, ν̅= ΔE hc in cm−1 . Here, h is Planck’s constant and c is the velocity of light in vacuum expressed in cm/s. Rotational spectrum is discussed in wavenumbers and hence we express energy in cm−1 . The energy in wavenumber units is, EJ = KE hc EJ = J(J + 1)h 8π 2Ic cm−1 This can be expressed as, EJ = BJ(J + 1) (J = 0, 1, 2, 3 ... ... ) where, B = h 8π 2Ic is called rotational constant.
University College of Science, Tumkur 4 To understand the pure rotational energy levels and spectra, determine energies of energy levels. i.e. For J = 0, E0 = 0, For J = 1, E1 = 2B cm−1 , For J = 2, E2 = 6B cm−1 , For J = 3, E3 = 12B cm−1 and so on. The pure rotational energy levels and pure rotational spectrum of diatomic molecules are represented as, Rotational energy levels Allowed transitions and rotational spectrum The allowed transition of molecule between the rotational energy levels is given by selection rules. For rotational energy levels the selection rule is given by, ΔJ = ±1 i.e. the transition for which J can change by ±1 are only allowed. The first spectral line (absorption of emission) in the rotational spectrum is due to transition of molecule from J = 0 to J = 1 level. The wavenumber of first line is ν1̅ = E1 − E0 = 2B cm−1 . Similarly, the second line is ν̅2 = E2 − E1 = 4B cm−1 , third line is ν̅3 = E3 − E2 = 6B cm−1 and so on. Therefore, the pure rotational spectra of diatomic molecule consist of spectral lines of wavenumbers 2B, 4B, 6B, 8B ... ... . . cm−1 . The spectral lines are equidistant and separated by 2B cm−1 . The pure rotational spectral lines lie in microwave region. Vibrational Spectra of molecules: The diatomic molecule vibrates about the mean position and therefore, the extension and compression of internuclear distance between the atoms of the molecule takes place due to vibrations. The minimum internuclear distance between the atoms of the molecules is called
University College of Science, Tumkur 5 bond length. The vibrational spectra are discussed assuming the vibrations of molecule as simple harmonic oscillations. Diatomic molecule vibrates and distance between the atoms changes with time. The vibration of the molecule can be compared with the vibration of masses connected by a spring. In the diatomic molecule two atoms of masses m1 and m2 vibrate as simple harmonic oscillator. The use of the “reduced mass” reduces the problem of two bodies into one body problem. We consider vibration of reduced mass about the centre of mass. Then frequency of oscillation of diatomic molecule is, νosc = 1 2π √ k μ where, k is the force constant and μ = m1m2 (m1+m2 ) is the reduced mass of the diatomic molecule. The quantized vibrational energy is given by, E = (v + 1 2 ) hνosc Joules where, h is Planck’s constant, v is the vibrational quantum number and v takes integer values, v = 0, 1, 2, 3 ... ... ... Vibrational spectra are discussed in wavenumber units. To convert energy in joules to cm−1 divide by hc, where, c velocity of light in vacuum expressed in cm/s. Therefore, the vibrational energy in wavenumber units is, Ev = E hc Ev = (v + 1 2 ) νosc c cm−1 Here, νosc c = 1 λ = ν̅, is the wavenumber. Therefore, Ev = (v + 1 2 ) ν̅ cm−1 (v = 0, 1, 2, 3 ... . . ) This expression gives the allowed quantized energies of vibrational energy levels. To understand the vibrational energy levels and spectra, determine energies of vibrational energies. i.e. For v = 0, E0 = 1 2 ν̅ cm−1 , For v = 1, E1 = 3 2 ν̅ cm−1 , For v = 2, E2 = 5 2 ν̅ cm−1 ,