Chemistry 2720 Fall 2000 Assignment 7 Solutions

1.

(a)

As in a hydrogen atom, the circumference must be a multiple of de Broglie wavelength:

$$2\pi r = n\lambda.$$

The wavelength is in turn related to the momentum by $\lambda = h/p$ and the momentum to the speed by p=mv. Thus we have

$$v = \frac{nh}{2\pi rm}.$$

The (kinetic) energy is

$$E_n = \frac{1}{2}mv^2 = \frac{n^2h^2}{8\pi^2r^2m}.$$

(b)

We need to calculate the energy of the n=3 to n=4 transition:

$$\begin{eqnarray*}% latex2html id marker 14 \Delta E & = & E_4-E_3 = \frac{h^2}{8... ...})}{2.18\times 10^{-18}\,\mathrm{J}}\\ & = & 91\,\mathrm{nm} \end{eqnarray*}$$

2.

The energies of the Li²⁺ spectral lines are given by

$$\Delta E = 9R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) ... ...17}\,\mathrm{J})\left(\frac{1}{n_f^2} -\frac{1}{n_i^2}\right).$$

Visible lines fall (roughly) in the range 2.5- $4.4\times 10^{-19}\,\mathrm{J}$. This means that 1/n_f²-1/n_i² must fall in the range 0.0128 to 0.0225. The n=5 to n=4 transition just falls in this range: $\Delta E_{5\rightarrow 4} = 4.4\times 10^{-19}$ so that $\lambda_{5\rightarrow 4} = hc/\Delta E_{5\rightarrow 4} = 450\,\mathrm{nm}$, which is a violet line. By trial and error, I have found the following set of lines:

ni	nf	$\lambda$ (nm)	color
5	4	450	violet
7	5	517	green
8	5	415	violet
9	6	656	yellow or orange
10	6	570	green

3.

We first compute $\Delta E$:

$$\begin{eqnarray*}\Delta E & = & \frac{hc}{\lambda}\\ & = & \frac{(6.626\,069\t... ...0^{-9}\,\mathrm{m}}\\ & = & 4.842\times 10^{-19}\,\mathrm{J}. \end{eqnarray*}$$

However

$$\Delta E = R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$$

so that

$$\frac{1}{n_f^2}-\frac{1}{n_i^2} = \frac{\Delta E}{R_H} = \fr... ...\,\mathrm{J}}{2.179\,874\times 10^{-18}\,\mathrm{J}} = 0.2222.$$

The trick now is to find a pair of values n_i and n_f which (at least approximately) satisfy the above relation. There is nothing for it but trial-and-error. After trying a few values, I found

$$\frac{1}{2^2} - \frac{1}{6^2} = 0.2222.$$

Therefore the line is due to an n=6 to n=2 transition.