Question

A $16nm$-long box has a thin partition that divides the box into a$4nm$-long section and a$12nm$-long section. An electron confined in the shorter section is in the $n=2$ state. The partition is briefly withdrawn, then reinserted, leaving the electron in the longer section of the box. What is the electron’s quantum state after the partition is back in place?

Short answer

The electron’s quantum state after the partition is back in place is$6$.

Step-by-step solution

Given Information

We have to given$16nm$a thin long box. It divides into two-part $4nm$and $12nm$.

Simplify

To solve this problem, we are going to use the fact that the energy of the electron will stay constant as it moves to the longer part of the box. However, the quantum state of the electron will be different therefore $L$is different now. Then, the energy of the $(n=2)$state in the first partition is equal to the energy of the $\left(n\right)$state in the second partition, and our goal is to find $n$. Since, we can write the following:

$E_2=E_n^{\text{'}}$

the prime indicates that this is the energy in the second partition. The energy of levels of an electron in a rigid box is given by:

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

Then,

$\begin{array}{rr}\frac{{\left(2\right)}^2{h}^2}{8m{\left(4\text{nm}\right)}^2}& =\frac{n^2{h}^2}{8m{\left(12\text{nm}\right)}^2}\\ \frac{{\left(2\right)}^2}{{\left(4\text{nm}\right)}^2}& =\frac{n^2}{{\left(12\text{nm}\right)}^2}\end{array}$

Calculating $n$:

$n=\sqrt{\frac{{\left(2\right)}^2\times {\left(12\text{nm}\right)}^2}{{\left(4\text{nm}\right)}^2}}=6$