Chem 6, 10 Section Spring Exam 2 Solutions

Transcription

1 Exam 2 Solutions 1. ( points) Dartmouth s FM radio station, WDCR, broadcasts by emitting from its antenna photons of frequency 99.3 MHz ( Hz). (a) What is the energy of a single WDCR photon? Photon energy and frequency are directly related: E = hν = J s s 1 = J (b) Suppose one of these photons strikes a hydrogen atom in a particular quantum state with a principal quantum number n. What is the minimum value for n that would allow this photon to ionize the atom? The state that can be ionized by this photon must have an energy equal to the negative of this photon energy, and we use the Bohr H atom energy expression to find the n value for that level. We write E n = J = J so that n = n J J = 5760 which is a really big quantum number! But you shouldn t have worried that it was too big because n can range all the way up to infinity and our FM photon had very little energy. (c) You now know n, but not l or m for this atom. Calculate the maximum average radius this atom could have, and in doing so, tell me what l and m values lead to this maximum value. (If you didn t get a value for n in part (b), assume n = 100, which isn t the answer to part (b).) If n is this huge number, then there are many, many possible l and m values for atoms with the same energy. But if we look at the expression for the average radius, we can see that as l increases from its minimum value of 0, the average radius decreases, which tells us that l = 0 has the largest radius, and if l = 0, then m = 0 as well: r nl = a 0 2 3n 2 l(l +1) = a 0 2 3(5756) 2 = m 2 3(5756) 2 = m That s one big H atom with a 2.63 mm radius!

2 2. ( points) Here are two questions about H atom wavefunctions. (a) An H atom starts in the state with the wavefunction on the left, then it emits a photon and ends in the state with the wavefunction on the right. What is the photon s wavelength? z z initial state final state Both of these orbitals are m = 0 (cylindrically symmetric) orbitals with no spherical nodes, as you can tell because the images have only one view and only one labeled axis, the z axis. The one on the left is a 3d z 2 orbital, and the one on the right is a 2p z orbital. Thus, or atom is undergoing a 3 2 transition, which is one you saw in lab in the Balmer series! It s the red emission. Its wavelength λ is given by so that ΔE = hc λ = E 3 E 2 = J λ = hc ΔE = hc J = m = 656 nm (b) One of the H atom s orbital radial distribution functions is shown below. What is its n quantum number, and how do you know this value for n?

3 n = 4 because: the orbital is an s orbital (l = 0), which means the only nodes it has are spherical nodes. We see the radial distribution function has three points where it is zero (other than at r = 0 and r = ), which tells us there are 3 spherical (radial) nodes, making n = = (3 points each) Which element or elements in the Periodic Table row from Li to Ne (a) has the smallest first ionization energy (b) has the smallest second ionization energy (c) has a negative electron affinity (d) has an excited configuration 1s 2 2s 2 2p 4 3s Li Be Be, N, Ne F (e) is the stable anion 1s 2 2s 2 2p 6 F (and/or O) (f) has a third ionization energy you could calculate exactly Li

4 (g) is spherical in its ground state (h) is isoelectronic to Na + when it is a neutral atom Li, Be, N, Ne Ne (i) has only one electron in an orbital with one nodal plane B (j) has two or more unpaired electrons in its ground state configuration C, N, O 4. ( points) The Li atom in its ground state can absorb many different photons, producing many different excited states. This question is concerned with the photon of longest wavelength that Li can absorb. This photon has an energy of J. (a) What is the electron configuration of the Li atom produced when ground state Li absorbs this photon? Just as in the Na atom you studied in lab and we discussed in class, the longest wavelength excitation (smallest energy excitation) promotes the s valence electron to the p orbital of the same principal quantum number. For Li, this means the 1s 2 2s 1 ground state configuration becomes the 1s 2 2p 1 excited state configuration. (b) The first ionization energy of Li is J. Draw an energy level diagram that locates the Li ground state, this excited state, and the state corresponding to Li + + e below. Label each of your energy levels. The ionization limit is at the zero of energy, and the ground state is J lower (i.e., at J). The excited state is above the ground state by an amount equal to the photon energy, placing it at J J = J. The diagram below summarizes these energies. 0 Li + + e 2 Energy/10 19 J Excited State Ground State

5 (c) Calculate the effective nuclear charge, Z eff, felt by the highest energy electron in this excited state. We note that the excited state valence electron has n = 2, and we equate the energy of this state, J, to the expression for the energy in terms of the effective nuclear charge: E n,l = J = ( J) Z 2 eff, n, l 2 2 (4) ( Z eff, n, l = 19 J) = 1.02 J 5. (4 points each) An electron is in a one-dimensional box of length L = 8.0 Å in the quantum state n = 6. (a) What is this electron s de Broglie wavelength? There are two ways (at least) to answer this question. The easy way says: picture the wavefunction. It fits three full sine waves in the box, and since there waves are de Broglie waves, the de Broglie wavelength is simply λ = L/3 = (8 Å)/3 = 2.67 Å. (Note that this is equivalent to saying that in general, L = nλ/2.) The other way, which is just fine, but more involved, says I know n, and thus I know the energy. The energy is all kinetic; so, I also know the momentum. If I know the momentum, I know the de Broglie wavelength. It s usually smart to tackle a string of calculations symbolically first rather than all numerically (this often saves time and reduces the chance for a calculation error somewhere). We can write E = h 2 n 2 8m e L 2 = p 2 2m e = (h/λ)2 2m e or h m e 8 Å 2 = h 2 simplifying gives Å 2 = 1 in agreement with the first method. λ 2 so that λ = 8 Å 3 2m e λ 2 = 2.67 Å (b) The electron gains energy via photon absorption and ends in the state with n = 8. Sketch the n = 8 wavefunction for this electron in the graph below. (Don t worry about the scale for the Ψ axis.) The n = 8 wavefunction must have 7 nodes. Here it is, showing that 4 full waves fit in the box:

6 0 x (c) Write (but do not try to evaluate) the expression for the probability that this n = 8 electron will be found between x = and x = 6 Å. Here, we need the integral of the square of the wavefunction over x from to 6 Å: Probability = 6 Å Ψ 2 (x) dx 6 Å = 2 8 Å sin 8 8 π Å x 2 dx = 1 6 Å sin 2 π x 1 Å dx 6. (3 + 4 points) The atmosphere around the sun contains lots of H and He atoms with some free electrons and hydride (H ) anions mixed in. The H anion is isoelectronic to He: 1s 2. We see the atoms through their atomic absorption spectra, but the hydride anions show up by absorbing light over a continuous range of wavelengths. (a) Given the electron affinity of H, J, calculate the longest wavelength of light that a hydride ion can absorb. Here, we set the electron affinity, EA, equal to the photon energy we seek, and express that photon energy in terms of the photon s wavelength: EA = hc λ or λ = hc EA = hc J = 1640 nm (b) What is the effective nuclear charge, Z eff, that the second 1s electron in H experiences? The electron affinity tells us the energy of the 1s orbital in H (it is EA) because our ionization limit energy is our energy zero and ionization of H means H H + e. Thus we can write

7 EA = J = J Z 2 eff so that Z eff = We see that the 1s electrons in H are doing a really good job of shielding the proton charge from each other! Note as well that Z eff is < 1.0, which is OK here because we re removing an electron and leaving behind a neutral atom, not a positive ion as happened in, for example, Problem 4 with Li atom being ionized.