91Ó°ÊÓ

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:

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

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 ${\mathrm{E}}_1:$

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

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

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