An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What is the length of the box?

The energy of an electron contained in a one-dimensional box have been observed to be 12 eV, 27 eV, and 48 eV at different periods.

We can begin by using the following expression for the quantum state of a particle in the box:

$\begin{array}{l}{\mathrm{E}}_{\mathrm{n}}={\mathrm{n}}^2{\mathrm{E}}_1\\ {\mathrm{E}}_{{\mathrm{n}}_1}:{\mathrm{E}}_{{\mathrm{n}}_2}:{\mathrm{E}}_{{\mathrm{n}}_3}=12:27:48\\ =4:9:16\text{(theratioofenergiesatdifferenttimes)}\\ {\mathrm{E}}_{{\mathrm{n}}_1}={\mathrm{n}}_1^2{\mathrm{E}}_1\left(\mathrm{First}\mathrm{particle}\right)\\ {\mathrm{E}}_{{\mathrm{n}}_2}={\mathrm{n}}_2^2{\mathrm{E}}_1\left(\text{secondparticle}\right)\\ {\mathrm{E}}_{\mathrm{n}3}={\mathrm{n}}_3^2{\mathrm{E}}_1\left(\text{thridparticle}\right)\\ {\mathrm{E}}_{{\mathrm{n}}_1}:{\mathrm{E}}_{{\mathrm{n}}_2}:{\mathrm{E}}_{{\mathrm{n}}_3}={\mathrm{n}}_1^2{\mathrm{E}}_1:{\mathrm{n}}_2^2{\mathrm{E}}_2:{\mathrm{n}}_3^2{\mathrm{E}}_3\left(\mathrm{Substitute}\mathrm{expressions}\mathrm{for}\mathrm{energies}\right)\end{array}$

We can now easily find ${\mathrm{n}}_1,{\mathrm{n}}_2$and ${\mathrm{n}}_3$from ratio.

$\begin{array}{l}{{\mathrm{n}}_1}^2=4\\ {\mathrm{n}}_2^2=9\\ {\mathrm{n}}_3^2=16\\ {\mathrm{n}}_1=\sqrt{4}=2\\ {\mathrm{n}}_2=\sqrt{9}=3\\ {\mathrm{n}}_3=\sqrt{16}=4\end{array}$

We can pick one of the three states and see what happens ${\mathrm{E}}_1:$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}1}={\mathrm{n}}_1^2{\mathrm{E}}_1 \\ 12\left(\mathrm{eV}\right)=2^2\cdot {\mathrm{E}}_1 \\ {\mathrm{E}}_1=\frac{12}{4}\left(\mathrm{eV}\right)\left(\mathrm{substitute}\right) \\ {\mathrm{E}}_1=3\mathrm{eV}\left(\mathrm{express}{\mathrm{E}}_1\right) \end{gathered}$$

We can now calculate the length of the box by using the following calculation for the particle's energy levels:

localid="1651145015345" role="math" $$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{h}}^2{\mathrm{n}}^2}{8{\mathrm{mL}}^2} \\ {\mathrm{E}}_1=\frac{{\displaystyle {\mathrm{h}}^2\cdot 1^2}}{{\displaystyle 8{\mathrm{mL}}^2}}\text{(substitute}\mathrm{n}=1\text{)} \\ \mathrm{L}=\sqrt{\frac{{\mathrm{h}}^2}{8{\mathrm{mE}}_1}}\left(\mathrm{express}\mathrm{L}\right) \\ \mathrm{L}=\sqrt{\frac{{\left(6.63\times {10}^{-34}\right)}^2}{\left(8\times 9\times 1\times {10}^{-31}\right)\times \left(3\times 1\times 6\times {10}^{-19}\right)}}\text{(substitute)} \\ \mathrm{L}=0.354\mathrm{nm} \end{gathered}$$