Problem Set #1 Solutions

1. We expect a linear relationship between the amounts of daughter and parent

which is confirmed by plotting the data.

The slope of the best fit line (0.001176) is equal to e^lt-1, which we solve for the age

a) t = 83 million years

The intercept yields the initial isotopic ratio

b)

which is significantly larger than that of modern basalts. This implies

c) that the granodiorites have a different source, with a different composition than the source of basalts. The granodiorite source must contain more ⁸⁷Sr, and hence more of the parent ⁸⁷Rb. A geologically plausible way to generate a source region with a different composition is to partially melt the mantle: the granodiorites are then the product of a second partial melting. This story holds together because Rb is an incompatible element: it is enriched in the partial melt, which then went on to freeze and finally to partially melt again to produce our granodiorites.

2. Periodic table exercise.

3. Electronic Configuration exercise. Example: Fe 1s²2s²2p⁶3s²3p⁶4s²3d⁶ or [Ar]4s²3d⁶.

4. Conservation of energy requires that the change in energy is equal to the energy of the photon that is absorbed to induce the transition

Recalling that the wavelength of light is related to its frequency by

we calculate the properties of the photons to be

a) n=5.38x10¹⁴, l=557 nm; n=7.34x10¹⁴, l=409 nm

which lie, respectively in the

b) yellow-green and violet

portions of the visible spectrum. Ruby then absorbs light at the mid- and high- frequency portions of the visible spectrum, but allows lower frequency (red) light to be transmitted, which explains its color.

Note that in detail we would have to consider other factors such as the partial transmission of colors other than red (e.g. blue sensu stricto), and the physiology of the eye as it pertains to the perception of color.

5. The frequencies of all three vibrational modes of the carbon dioxide molecule lie in the

a) infrared portion of the spectrum.

b) The infrared activity is as follows:

The symmetric stretch is inactive. The reason is that the molecule has no dipole moment in its equilibrium configuration and by stretching it symmetrically, no polarity is introduced.

The asymmetric stretch is active since vibration in this way induces a dipole moment: the molecule is briefly polar as it vibrates.

The bend is also active since it briefly produces a non-zero dipole moment. The molecule is briefly bent, much like a water molecule.

c) The asymmetric stretch and the bend absorb part of the infrared light emitted by the earth's surface, thus trapping this energy and contributing to the blanketing effect of the atmosphere.

d) Oxygen and Nitrogen are not greenhouse gases because they have no infrared active vibrational modes. Each has a single mode (stretch) which does not induce a dipole moment.

6. The shape of the charge density depends on the charge of the nucleus a)

The curves for smaller values of Z are shown for comparison.

b) The shape of the charge density depends on the charge of the nucleus. Notice that for large values of Z, the charge density is more concentrated around the nucleus. The reason is that the larger nucleus attracts the electron more strongly through the Coulomb force. Also, for larger values of Z, the charge density falls off more rapidly towards zero. This is also because the larger nucleus concentrates the charge more towards smaller values of distance.

c) Li⁺ and Be²⁺ are closed shell second-row ions. This means that their outermost shell is a completely filled 1s shell. We can therefore use our graph to estimate the radii. We wish to identify a radius that contains most of the charge. The charge density approaches zero asymptotically, never quite reaching it. It is therefore impossible to uniquely define the radius of an atom or ion. We can make a reasonable estimate however, by assigning the point at which the charge density becomes very small (say less than 0.1) to be the radius. Using this criterion, we find that the radii of the ions are:

Li⁺ 1.23 Bohr or 0.651 Å.

Be²⁺ 0.97 Bohr or 0.513 Å.

d) which are comparable to the Shannon and Prewitt radii for 6-fold coordination: 0.74 Å and 0.45 Å, respectively. In particular, their relative magnitudes are correct. This makes sense because it is the charge density of an ion that largely determines its size. The more spread out the charge density of an ion is, the larger will be the repulsive forces between it and neighboring ions at a given distance. In detail, the Shannon and Prewitt radii differ from our estimates because they are based on different criteria, which are no better or worse, and no more or less correct.