91影视

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

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 $n$:

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