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HydrogenAtomProblemSetKey

HydrogenAtomProblemSetKey - H drogen Atom Problem Set...

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Unformatted text preview: H drogen Atom Problem Set Kathleen Kristian em 1403 Fall 2008 1. On the axes below, draw and label the energy levels for (A) the 1D particle in a box and (B) the hydrogen atom. Include at least 5 energy levels in your sketch. Label E1 and Em with their actual numerical values for both (A) and (B). Remember units! (A) flflts (B) €04: ~’ -"" V\:00 Zero '5 A i \J n ' Li \J “:5 § E ~ x; 8 g “t m m In: 3, V\=3 Mia h=9~ 5, :: he; 5.4mm“ E1 = ’ * “3‘ M La any WWI «gnCNMEmzoo A few things to notice: -- The energy levels for a particle in a box get farther apart as 11 increases, while the energy levels for the hydrogen atom get closer together as 11 increases. -- All of the possible energies for an electron in a hydrogen atom are negative numbers, except for the maximum energy, Em, which is equal to zero. All of the possible energies for a particle in a 1-D box are positive, and Ego is equal to infinity. -- The minimum possible energy for both situations is NOT equal to zero. Hydrogen Atom Problem Set Kathleen Kristian Chem 1403 Fall 2008 2. Which quantum number(s) are necessary to specify the electronic state of a one- electron atom or ion? To specify the electronic state of a one-electron atom or ion you need to know 11, l, and m,. 3. Which quantum number(s) are necessary to determine the energy of a one-electron atom or ion? The energy of a one-electron atom or ion is determined only by n, the principal quantum number (recall that E, = —13.6(Zz/ n2) eV). 4. How many d orbitals are there in the nth energy level of the hydrogen atom? How many f orbitals are there in the nth energy level of the hydrogen atom? The d orbitals, regardless of what shell they are in (that is, regardless of what value of 11 they have), must have I = 2. This means that the possible values for m, are 0, =1, :2. Because there are five possible values of m, there are five distinct d orbitals. Likewise, the f orbitals must have I = 3. Therefore the possible values of m, are 0, =1, :2, :3. Because there are seven possible values of m,, there are seven distinct d orbitals. The number of m, values for a given value of l is fixed and does not depend on n. In fact, the number of possible m, values is equal to 21 + 1, and is equal to the number of orbitals with that specific value of l. 5. (a) Radial distribution functions (12‘1“) for two different H atom orbitals are shown below. Which is higher in energy? Explain why in one sentence or less. The orbital represented by (B) is higher in energy because it has more nodes. As orbital energy increases, the number of nodes also increases. (b) On the plotsidiraw an arrow pointing to each node. (A) (B) «c N a: N5“ k \hov'l r r _ — _ Hydrogen Atom Problem Set Kathleen Kristian Chem 1403 Fall 2008 (c) Is there a node at the origin in this plot? Explain. Based solely on the plots above, without knowing what specific wavefuncfions with what specific quantum numbers they correspond to, you cannot determine whether there is a node at the origin. The function that is plotted is 1211’2, and at the origin, r = O, which makes this function equal to zero regardless of what the value of II‘2 is. It could be zero or any other number. In actuality, there is not a node at the origin in these plots. (A) is a plot of the radial distribution function for a 25 orbital and (B) is the radial distribution function for a 35 orbital. If you look at the actual mathematical functions ‘112 for these two states (in your textbook), you will see that at r = O, the function ‘1”2 takes a nonzero value. (d) In class you learned that the square of the wavefunction 0112) is equal to the probability density. Why do you think we make plots of r2 1P2 vs. r instead of just ‘112 vs r? First, what is the radial distribution function and how did we find it? The radial distribution function represents the probability of finding the electron at a given distance r from the nucleus—that is, it is the probability of finding the electron at any point in a spherical shell” with an inner radius equal to r and an outer radius equal r + Ar (for those with calculus, it’ s r + dr): ( 7,, I> ‘V’\ (x063 W Ar: Hm‘clihtss 010 Mo. shell. IQ‘PS Va "Wig than M Mme, ’1’: ‘HML shell 511’ Sung“; area 0? 3 M w] ram: €06,331 JFD T‘. If you allow the distance from r to r + Ar to be infinitesimally small, then the probability of finding the electron inside the spherical shell is approximately equal to the probability of finding the electron anywhere on the surface of a sphere with radius r. Hydrogen Atom Problem Set Kathleen Kristian Chem 1403 Fall 2008 How do you figure out what the function actually is? We calculate the probability density at every point in that spherical shell, sum those probability densities, and multiply by the tiny, tiny volume of the spherical shell. For those of you with calculus, we integrate the probability density | 1Mr, 0, (1)) | 2 with respect to r, 0, and :1), making the limits on the dr integral go from r to r + dr. This calculation shows that the radial distribution function is equal to R(r)2rz, where R(r) is the part of the wavefunction ‘11 that depends on only r (not 0 or ¢). For any 5 orbital (state with l = 0), the wavefunction 11’ only depends only on r in the first place, so the radial distribution function R(r)2r2 is equal to 1P2r2 (and that is why it is written that way in the pictures above). In your book, the radial disu‘ibution function is given as 4::r2R2 vs r. The extra constants at the beginning are no big deal—you can write the function with or without the constants, depending on how you’ve defined R(r), etc (and hopefully you can see that multiplying every value of R(r) by a constant Will not make the plot look any different qualitatively). Notice that these plots of the radial distribution function for the 25 and 35 orbitals have maxima and minima, though the wavefunction ‘11 is just a decaying exponential for these states. The curves are shaped the way they are shaped because we are multiplying by the factor r. The variable r starts out equal to zero and increases. A decaying exponential starts out at some finite number and ends up approaching zero. So these two functions are fighting with each other as r goes from zero to infinity, and you end up with a plot that has maxima. Qualitatively, think about the series of spheres with radius r, where r is varying from zero to infinity. The spheres with very small radii have very small surfaces, and the spheres with large radii have large surfaces. So even though the wavefunction ‘P (and therefore the probability density l1’2) is very large at small values of 1', there just aren’t as many places in space for the electron to be when r is small. So the radial distribufion function takes into account the number of possible positions for the electron at given radius r, and that is why we use it. Note: This question is here to make you think about the radial distribution function and what its purpose is. I did not expect you to come up with an answer as detailed as the one above. Hydrogen Atom Problem Set Kathleen Kristian Chem 1403 Fall 2008 6. On the axes below, draw the indicated orbitals. Indicate phase using either shading or + and — signs. List the values 11, l, and m (where appropriate) for each orbital. <a>1s (b) 2p, n z a 1 y fl :1 no m1 Value Um be, Steak»! (c0217, 1‘: i «(A plant 1,5 & nuke, 7. What is the energy of the 25 orbital in a hydrogen atom? What is the energy of the 2pz orbital in a hydrogen atom? In a hydrogen or hydrogen-like atom, all of the states with the same value of n have the same energy. The 23 and the 2px orbitals have the same energy, and it is E2 = —13.6 eV (12/ 22) = —3.4 eV. These two states are degenerate in a hydrogen atom. We will see that this is not the case for atoms that have more than one electron. ...
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