Chemistry 2720 Fall 2000 Test 2 Solutions

1.

(a)

d corresponds to $\ell=2$, so the possible values of $m_\ell$ are -2,-1,0,1,2. Thus, there are five of these.

(b)

For n=3, $\ell$ can't be any larger than 2. However, f orbitals correspond to $\ell=3$ so there are in fact no 3f orbitals.

2.

There are many examples:

• Photoelectric effect: Due to the relationship between the energy and frequency of photons, light of too low a frequency cannot cause the ejection of electrons from metals. The intensity (brightness) of the light is relevant only insofar as additional photons can cause more ejections, but only if the individual photons have enough energy.
• Electron/neutron diffraction: Particles (such as electrons) have wave properties so that they can diffract through matter.
• Spectroscopy: Quantum systems generally have discrete energy levels. As a result, they can absorb or emit radiation only at certain wavelengths.
• Tunneling: Quantum mechanical particles can be transmitted through energy barriers which, classically, they would not be able to surpass.

3.

$$\begin{eqnarray*}p & = & mv = (1.675\times 10^{-27}\,\mathrm{kg})(8000\,\mathrm... ...10^{-23}\,\mathrm{kg\,m/s}} = 4.945\times 10^{-11}\,\mathrm{m} \end{eqnarray*}$$

4.

This is a straightforward application of Bragg's law. Since we get more intensity from the first order of diffraction, this is the one we should go after.

$$\begin{eqnarray*}% latex2html id marker 28 \sin\theta & = & \frac{n\lambda}{2d} ... ...hrm{pm})}\\ & = & 0.0879\\ \therefore\theta & = & 5.0^\circ \end{eqnarray*}$$

5.

(a)

We start from the uncertainty principle

$$\Delta x\Delta p\ge \frac{h}{4\pi}$$

and note that

E = cp

for massless particles. Therefore

$$\begin{eqnarray*}% latex2html id marker 39 \Delta E & = & c\Delta p\\ \therefo... ...h}{4\pi}\\ \therefore\Delta x\Delta E & \ge & \frac{hc}{4\pi} \end{eqnarray*}$$

(b)

Using the inequality derived above, we have

$$\begin{eqnarray*}% latex2html id marker 47 \Delta E & \ge & \frac{hc}{4\pi\Delta... ...{-34}\,\mathrm{J/Hz}}\\ & = & 2.39\times 10^{16}\,\mathrm{Hz} \end{eqnarray*}$$

This is enormous, considering that optical frequencies are in the range of $10^{14}\,\mathrm{Hz}$.

6.

The ionization energy of a hydrogenic atom is

E_I = -E₁ = Z²R_H.

Our task is to solve this equation for Z. First, we must convert the ionization energy to Joules:

$$E_I = (24.587\,\mathrm{eV})(1.602\,176\times 10^{-19}\,\mathrm{J/eV}) = 3.9393\times 10^{-18}\,\mathrm{J}$$

Thus we have

$$Z_\mathrm{eff} = \sqrt{\frac{E_I}{R_H}} = \sqrt{\frac{3.9393\... ...}\,\mathrm{J}}{2.179\,874\times 10^{-18}\,\mathrm{J}}} = 1.34.$$

This value is less than 2 because the electron which remains after ionization shields the removed electron from the full nuclear charge.

7.

The first line of the Pfund series corresponds to the n=6 to n=5 transition. The energy of this line is

$$\begin{eqnarray*}\Delta E_{6\rightarrow 5} & = & R_H\left(\frac{1}{5^2}-\frac{1}... ...10^{-18}\,\mathrm{J})\\ & = & 2.66\times 10^{-20}\,\mathrm{J} \end{eqnarray*}$$

The frequency is

$$\nu = \frac{\Delta E_{6\rightarrow 5}}{h} = \frac{2.66\times ... ...mes 10^{-34}\,\mathrm{J/Hz}} = 4.02\times 10^{13}\,\mathrm{Hz}$$

The corresponding wavelength is

$$\lambda = \frac{c}{\nu} = \frac{2.997\,925\times 10^8\,\mathrm{m/s}}{4.02\times 10^{13}\,\mathrm{Hz}} = 7.46\,\mu\mathrm{m}$$

The others are computed analogously. Here are the results:

ni	$\lambda (\mu\mathrm{m})$
6	7.46
7	4.65
8	3.74
9	3.30

8.

(a)

We first want to calculate the minimum (kinetic) energy from the particle-in-a-box formula:

$$E_{K\mathrm{(min)}} = E_1 = \frac{h^2}{8mL^2}$$

We need the mass of a single protein molecule:

$$m = \frac{28\,\mathrm{kg/mol}}{6.022\,142\times 10^{23}\,\mathrm{mol}^{-1}} = 4.65\times 10^{-23}\,\mathrm{kg}$$

Thus,

$$E_{K\mathrm{(min)}} = \frac{(6.626\,069\times 10^{-34}\,\math... ...imes 10^{-3}\,\mathrm{m})^2} = 5.25\times 10^{-40}\,\mathrm{J}$$

We use $E_K = \frac{1}{2}mv^2$ to calculate a minimum speed:

$$\begin{eqnarray*}v_\mathrm{min} & = & \sqrt{2E_{K\mathrm{(min)}}/m} = \sqrt{2(5.... ...{-23}\,\mathrm{kg})}\\ & = & 4.75\times 10^{-9}\,\mathrm{m/s} \end{eqnarray*}$$

(b)

The average kinetic energy at room temperature is

$$\bar{E}_K = \frac{3}{2}RT = \frac{3}{2}(8.314\,510\,\mathrm{J\,K^{-1}mol^{-1}})(293\,\mathrm{K}) = 3654\,\mathrm{J/mol}$$

On a per molecule basis, this is

$$\bar{E}_K = \frac{3654\,\mathrm{J/mol}}{6.022\,142\times 10^{23}\,\mathrm{mol}^{-1}} = 6.07\times 10^{-21}\,\mathrm{J}$$

This is enormously larger than the minimum imposed by quantum mechanics, so quantum effects are of no particular significance in this system.