Homework Week 1. electronic excited state. electronic ground state

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1 Statistical Molecular Thermodynamics University of Minnesota Homework Week. The diagram below shows the energy of a diatomic molecule as a function of the separation between the atoms, R. A electronic excited state energy,v(r) B C electronic ground state D E R Which dashed line marks the defined zero, V (R) = 0, for the electronic energy of the diatomic? (a) A (b) B (c) C (d) D (e) E The zero of electronic energy for a diatomic molecule is taken to be that of the separated constituent atoms at infinite distance from one another. Thus, the correct answer is (C). t is not (E); this represents the minimum (not zero) value of the electronic energy. 2. f a hydrogen atom in the first excited state (n = 2) relaxes to the ground state (n = ) it will emit a photon to conserve energy. Which of the following is closest to the wavelength (λ) of the emitted photon?

2 (a) 75 nm (b) 22 nm (c) 225 nm (d) 500 nm (e) 667 nm (f) 000 nm ε = ε n=2 ε n= = ( cm ) ( 2 ) 2 2 ε = ( cm ) ( ) = ( cm ) ( ) 3 = cm Convert to wavelength: λ = ν = cm = cm ( ) 0 λ = nm cm = 22 nm cm 3. You have a diatomic molecule. You measure the bond dissociation energy and find that it is D 0 = 400 kj mol. You measure the vibrational frequency using R absorption and find that there is an absorption line at ν = 2000 cm. What is the electronic energy, D e, of the bond in this molecule? (a) kj mol (b) kj mol (c) kj mol (d) 42.0 kj mol (e) kj mol (f) kj mol Start with D e = D 0 + hν. Remember to convert the second term to kj/mol, 2 hν = N A hc ν: ( D e = J mol mol ) ( J s) ( cm s ) (2000 cm ) + 2 D e = J mol J mol D e = J mol = 42.0 kj mol 4. Convert ν = 000 cm to kj mol using the following values for standard constants: h = J s, c = m s, N A = mol.

3 (a) kj mol (b) kj mol (c) kj mol (d).96 kj mol (e).96 kj mol (f) 08.4 kj mol (g) 080 kj mol f Planck s constant is taken as h = J s, then E = hν (per molecule) and E = N A hν (per mole, where N A is Avogadro s number) and, ν = c ν, so E = N A hc ν = ( mol ) ( J s) ( cm s ) (0 3 cm ) = J mol =.96 kj mol 5. A ground-state hydrogen atom absorbs a photon of light that has a wavelength of 97.2 nm. t then gives off a photon that has a wavelength of 486 nm. What is the final state of the hydrogen atom (n =?). (a) (b) 2 (c) 3 (d) 4 (e) 2.3 (f) 8 (g) 5 For this problem start by making a diagram of the relevant energy levels, see figure. First we realize from the diagram that the energy difference between the initial (ground state) and the final state of our hydrogen atom is equal to the energy difference between the absorbed photon and the emitted photon, E final E ground = E photon absorbed E photon emitted To calculate the energy difference we will need to convert wavelength to an energy unit, and we can do this combining E = hν and c = νλ to give us E = h c λ, c E final E ground = h h λ ( absorbed = hc λ absorbed c λ emitted λ emitted ) = ( J s ) ( cm s ) ( cm cm = J

4 E intermediate Energy photon absorbed λ = 97.2 nm Eabsorbed Eemitted λ = 486 nm photon emitted Efinal Eground E final E ground n final =? n = Figure : Hydrogen atom energy levels for problem 3. Now that we know the energy difference between the ground state and the unknown final state, we can use the equation from question 2 to express the energy difference in terms of n. We know that n ground = for the ground state, which only leaves us to solve for n final, J n 2 final ( J ) ( E final E ground = J J = J n 2 final J n final = n 2 ground J J n 2 ground ) ( ) n 2 final = n 2 final n final = 2 = J = J = J J = 0.25 = 4 6. Given that ν = 2330 cm and D 0 = 7875 cm for N 2 (g), calculate the value of D e. Note that the value of ν is the position of an absorption peak in the R/Raman spectrum of N 2. This means that it is equal to the gap between quantized vibrational levels. (Hint: See the lecture slide titled Dissociation Energy) (a) cm (b) 8045 cm (c) cm (d) 7875 cm

5 (e) cm (f) cm The electronic energy, D e, is the sum of the bond dissociation energy, D 0, and the zero point vibrational energy, hν, 2 D e = D hν The only tricky issue is keeping the units all the same in energy. Since they were given in the same units it is trivial to calculate D e in wavenumbers. The only other thing to note is that we were given the position of the R/Raman absorption ( ν = 2330 cm ). Since this is the gap between two adjacent quantized vibrational levels, we know that it is equal to hν in energy. As a result, we need to multiply this energy by in the 2 last term, ( ) D e = 7875 cm cm = cm 2 This can, of course, be converted into other units, for example to get kj/mol, E = N A hν = N A hc ν D e = 956 kj mol 7. The energy difference between the J = 0 and J = rotational levels of carbon monoxide ( 2 C 6 O(g)) is ν = MHz. Calculate the energy difference between J = 0 and J = 2 and give the answer in cm. (a) cm (b) cm (c).54 cm (d) cm (e) cm (f) cm Here we start with the energy levels of the rigid rotor, E J = 2 J (J + ) 2 and consider the energy difference between two adjacent levels, J + and J, E J+ E J = 2 2 E J,J+ = 2 2 (J + ) ((J + ) + ) J (J + ) 2 = 2 [(J + ) (J + 2) J (J + )] 2 = 2 [ J 2 + 3J + 2 J 2 J ] 2 = 2 [2 (J + )] 2 (J + )

6 Now write the energy difference between J = 2 and J = 0 as the sum of the difference between J = 0 and J =, and J = and J = 2, E J=0,J=2 = E J=0,J= + E J=,J=2 = 2 = 3 2 (0 + ) + 2 ( + ) Compare this with just E J=0,J= (which was given in the problem), and you find that, E J=0,J= = 2 (0 + ) = 2 E J=0,J=2 = 3 E J=0,J= Using the given value for E J=0,J=, E J=0,J=2 = 3 ( MHz ) ( ) 0 6 Hz MHz Convert this to wavenumbers, = Hz ν = ν c = s cm s =.54 cm 8. Calculate the degeneracy of the first 4 rotational levels of carbon monoxide: (a) 0,,3,5 (b) 0,2,4,6 (c) (,),(3,3) (d),3,6,9 (e) 2,3,5,8 (f),3,5,7 (g) Cannot be determined, there are only three rotational levels possible. (h),2,3,4 The degeneracy is g J = 2J +, J g J = 2J + 0 g 0 = 2 (0) + = g = 2 () + = 3 2 g 2 = 2 (2) + = 5 3 g 3 = 2 (3) + = 7

7 9. Determine the number of various (translational, rotational, vibrational) degrees of freedom of N 2 : (a) 3,2, (b) 6,2, (c) 3,3,0 (d) 3,,2 (e) 5,3, (f) 6,, (g) 6,2,2 The total number of degrees of freedom must be 3 times the total number of atoms (3N), since each atom can move in 3D space. There are always 3 translational degrees of freedom (the whole molecule moving in 3D space). For rotations, the only question is whether the molecule is linear or nonlinear. For a linear molecule there are 2 rotations, and for a nonlinear molecule there are 3 rotations. The vibrations are whatever is left from the total (total minus translations minus rotations). Thus, for a linear molecule there are 3N 5 vibrations, and for a nonlinear molecule there are 3N 6 vibrations. Nitrogen, being diatomic, is perforce linear. molecule N total (3N) translations rotations vibrations N Determine the number of various (translation, rotation, vibration) degrees of freedom of C 2 H 6 : (a) 3,3,5 (b) 3,3,8 (c) 8,3,3 (d) 5,3,8 (e) 3,3,24 (f) 24,3,3 (g) 3,2,9 (h) 3,2,7 The total number of degrees of freedom must be 3 times the total number of atoms (3N), since each atom can move in 3D space. There are always 3 translational degrees of freedom (the whole molecule moving in 3D space). For rotations, the only question is whether the molecule is linear or nonlinear. For a linear molecule there are 2 rotations, and for a nonlinear molecule there are 3 rotations. The vibrations are whatever is left from the total (total minus translations minus rotations). Thus, for a linear molecule there are 3N 5 vibrations, and for a nonlinear molecule there are 3N 6 vibrations. Ethane (C 2 H 6 ) is non-linear.

8 molecule N total (3N) translations rotations vibrations C 2 H