Question

FIGURE EX $40.5$is the probability density for an electron in a rigid box. What is the electron’s energy, in $eV$?

Short answer

The electron’s energy, in $eV$ is$16.9eV$.

Step-by-step solution

Given Information

We need to find the electron’s energy, in$eV$.

Simplify

The given graph of the probability density has three antinodes, which means that the wave function also has three antinodes. And we know that the graph of ${\psi }_n\left(x\right)$has $\left(n\right)$antinodes .). So, the wave function shown in the graph corresponds to the $n=3$state. But the energy of an electron in a rigid box is given by:

$E_n=n^2\frac{h^2}{8m{L}^2}$

Notice that $L$ is given to be$\left(0.45nm\right)$ from the graph, which is the maximum value of $x$

substitute the numerical values for the quantities in the equation above to find $E_3$

$E_3=\frac{9\times {\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\text{kg}\right)\times {\left(0.45\times {10}^{-9}\text{m}\right)}^2}=2.7\times {10}^{-18}\text{J}$

or in electronvolt

$\begin{array}{c}{E}_3=2.7\times {10}^{-18}\text{J}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\\ E_3=16.9\text{eV}\end{array}$