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Chemistry 2720 Fall 2000 Assignment 7 Solutions

1.
(a)
As in a hydrogen atom, the circumference must be a multiple of de Broglie wavelength:

\begin{displaymath}2\pi r = n\lambda.\end{displaymath}

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

\begin{displaymath}v = \frac{nh}{2\pi rm}.\end{displaymath}

The (kinetic) energy is

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

(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

\begin{displaymath}\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).\end{displaymath}

Visible lines fall (roughly) in the range 2.5- $4.4\times
10^{-19}\,\mathrm{J}$. This means that 1/nf2-1/ni2 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

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

so that

\begin{displaymath}\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.\end{displaymath}

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

\begin{displaymath}\frac{1}{2^2} - \frac{1}{6^2} = 0.2222.\end{displaymath}

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

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Marc Roussel
2000-11-14