CHEM 301: Homework assignment #5

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1 CHEM 30: Homework assignment #5 Solutions. A point mass rotates in a circle with l =. Calculate the magnitude of its angular momentum and all possible projections of the angular momentum on the z-axis. 0% The magnitude of the angular momentum is J = h l l + = J s + = J s. The projection of angular momentum onto the z-axis is determined by the quantum number m l. For l =, there are three possible values of m l :, 0, and +. The projections of the angular momentum for these three cases are: J z = hm l = m l = + : h; m l = 0 : 0; m l = : h. Schematically draw all angular momentum cones consistent with l =. 0% See Figure. Figure : The red cone corresponds to m l = +, the green cone actually, a circle in the xy-plane to m l = 0, and the blue cone to m l =. The length of the angular momentum vector J is J = h l l + = h since l = and is the same for all three cones. The projection of the angular momentum vector J onto the z-axis is J z = hm l and is the same for all vectors within the same cone. The angle θ is measured from the positive direction of the z-axis. Calculate the half-angles for each of the cones. 0% The angle θ is measured from the positive direction of the z-axis. For m l = +, its cosine can be written as: cos θ ml =+ = J m l=+ z J = h h =.

2 Thus, θ ml =+ = arccos =. It is evident that the cone for m l = is just a mirror reflection in the xy-plane of the cone for m l = +. Thus, it lies at an angle of, as measured from the negative direction of the z-axis, or θ ml = = = 3 as measured from the positive direction of the z-axis. Finally, for m l = 0 the angular momentum J is perpendicular to the z-axis since its projection onto the z-axis is J m l=0 z = 0. Thus, θ ml =0 =.. The force constant for the HI bond is k = 3 N/m. Calculate the vibrational frequency of an HI molecule. 0% Approximating the HI molecule by a harmonic oscillator, we can calculate its vibrational frequency using the formula ν = k µ, where µ is the reduced mass of the molecule: µ = m Hm I.008 a.u. 6.9 a.u. = m H + m I.008 a.u a.u. =.000 a.u. = kg and k = 3 N/m is the force constant. Thus, 3 N/m ν = kg = s. What is the average potential energy in the ground state of this molecule? 0% To calculate the average value s of any quantum mechanical observable s we can use one of the postulates of quantum mechanics: s = ψ x Ŝψ x dx, where Ŝ is the operator that corresponds to observable s and ψ x is the wavefunction of the system studied. We need to find the average potential energy in the vibrational ground state of a molecule that we describe as a harmonic oscillator. The potential energy operator is just the potential function V x, for the harmonic oscillator it is: ˆV = V x = kx.

3 From the lecture we know that the eigenfunctions for the harmonic oscillator are normalized products of Hermite polynomials and Gaussian functions: ψ v x = α v v! normalization constant H v α x Hermite polynomial exp αx. Gaussian function We are interested in the vibrational ground state, where v = 0: α ψ 0 x = H 0 0! 0 α x exp αx = = see lecture 5, slide = = α exp αx. Since ψ 0 x is a real-valued function, ψ 0 x = ψ 0 x. Plugging in the wavefunction in the ground state of a harmonic oscillator and the potential energy operator for the harmonic oscillator into the expression for the average value, we get: V = = k kx ψ0 x ψ 0 x dx = k x α exp αx dx = k α x ψ 0 x dx = = [look up the integral in a table of integrals] = k = k α. x exp αx dx = α α α = Here, kµ α = h. Plugging this into the expression for the average potential energy that we obtained results in: V = k α = k h km = h k m = h k m } {{} =ν = hν. The average potential energy is thus one half of the zero-point energy: E 0 = hν. Plugging in the numbers, we get: V = hν = J s s = J = 7 mev. 3

4 What is the wavelength that would excite this molecule into vibration? What part of the electromagnetic spectrum does this wavelength correspond to? 0% The energy levels of a harmonic oscillator are given by E = hν v +. In the context of this problem, to excite a molecule into vibration means to take it from the vibrational ground state v = 0 to the first vibrational excited state v =. The energy difference for these two states is: E = hν + hν 0 + = hν. This is the energy that a photon must supply to the molecule to facilitate this transition. The energy of a photon can also be written as E photon = hν photon. By comparing the expressions for E and E photon, we see that the frequency of the photon, ν photon, must be the same as the characteristic frequency of the harmonic oscillator, ν, that we calculated earlier. To convert the frequency to the wavelength, λ, use the relation c = λν, where c is the speed of light. Thus, λ = c ν = m s s = m =.33 µm. How would the vibrational frequency change if hydrogen were replaced by deuterium? 0% Since deuterium is approximately twice as heavy as hydrogen, but is still much lighter than iodine, the reduced mass for the DI molecule would be approximately a.u. instead of a.u. Since the frequency ν = k µ, then, assuming that k remains the same, increasing µ by roughly a factor of would reduce ν by roughly a factor of. 3. What manifestations of Heisenberg uncertainty principle have we seen in the lecture about the particle on a sphere and the harmonic oscillator? Give two examples, with brief explanations. 30%

5 Heisenberg s uncertainty principle says that if the operators Ŝ and ˆQ that correspond to observables s and q, respectively, commute, then s and q can be known simultaneously to arbitrary precision. However, if Ŝ and ˆQ do not commute then the uncertainties in the values of s and q, s and q, respectively, must satisfy s q h. For a particle on a sphere we have seen that although the angular momentum J and one of its projections can be known at the same time to arbitrary precision. However, once one projection of the angular momentum usually denoted J z is known, the other two projections usually denoted J x and J y must be completely indefinite. We can expect this to be a consequence of Heisenberg s uncertainty principle and may predict that the operator of the angular momentum projection Ĵz commutes with the operator of angular momentum Ĵ, but Ĵz does not commute with the operators of the two other projections of angular momentum, Ĵx and Ĵy. We have not written these operators out explicitly, but this, indeed, turns out to be the case. Similarly, both Ĵx and Ĵy will commute with Ĵ, but not with each other or with Ĵz. For a particle in a box we have seen that the existence of zero-point energy is a consequence of the position-momentum uncertainty principle a specific manifestation of Heisenberg s uncertainty principle, where the observable s is position x, and the observable q is momentum p. Let s recap that argument. Since a particle can only be found inside the box, its position x is not completely indefinite, and hence the uncertainty in position x. Since the operators of position and momentum do not commute, the uncertainty relation x p h must be satisfied. For x, this condition can only be satisfied if p 0. However, if p 0, we don t know exactly what the value of p is. We don t know what the value of p is exactly, but we definitely cannot say that p = 0 if this were the case, we would be certain about the value of p, and the uncertainty p would be 0. But if p 0, the kinetic energy E k = p m 0. Thus, the position-momentum uncertainty principle requires that there be a non-zero minimum energy that a particle that is confined to a box must possess: the zero-point energy. Zero-point energy also exists for the harmonic operator, and we can expect it to also be the consequence of the position-momentum uncertainty principle. Similar reasoning to the case of a particle in a box is possible. Although a particle can penetrate into the walls of a harmonic potential, it is still most likely to be found either inside the harmonic potential or just 5

6 outside its walls. Thus, the position of the particle is not completely indefinite: x. The position-momentum uncertainty principle requires that x p h, so if x, then p 0. Thus, we can t say that p or the kinetic energy that depends on it are exactly zero. There is a a non-zero minimum energy that a particle in a harmonic potential must have, the zero-point energy. 6