91影视

The side length of 3D well is L.

The number of identical particles is n = 15 .

The mass of the identical particles is m.

The lowest energy wave function has $\left({\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}\right)\mathbf{=}\left(\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\right)$. The next two lowest have (2,1,1), (1,2,1) and (1,1,2) . By combining each from these two there will six pairs for adding in existing two. Next higher will be (2,2,1) ,(2,1,2) and (1,2,2). This adds six to existing eight states. The fifteenth state is(3,1,1) ,(1,3,1) and (1,1,3) . for the last particle

The lowest possible total energy is given as:

$E_t=\left(鈭�\left(n_x^2+n_y^2+n_z^2\right)\right)\left(\frac{{蟺}^2{h}^2}{2m{L}^2}\right)$

Here, h is the Planck鈥檚 constant.

Substitute all the values in the above equation.

$$\begin{gathered} E_t=\left[2\left(1^2+1^2+1^2\right)+6\left(2^2+1^2+1^2\right)+6\left(2^2+2^2+1^2\right)+\left(3^2+1^2+1^2\right)\right]\left(\frac{{蟺}^2{h}^2}{2m{L}^2}\right)\backslash \mathrm{hfill} \\ {E}_t=\frac{107{蟺}^2{h}^2}{2m{L}^2} \end{gathered}$$

Therefore, the lowest possible total energy is $\frac{107{蟺}^2{h}^2}{2m{L}^2}$.

Two among the three quantum numbers are unity, which gives maximum wave function at the centre of 3D well. The other quantum number is three for which there will be three maxima. This indicates that particle is most likely found at three points. The probability of finding particle is at the centre of the 3D well.

Therefore, the highest energy particle is most likely to found at the centre of the well.