If you double the width of a one-dimensional infinite potential well, (a) is the energy of the ground state of the trapped electron multiplied by$\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{,}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}$ or some other number? (b) Are the energies of the higher energy states multiplied by this factor or by some other factor, depending on their quantum number?

The length of the potential well is double.

The expression to calculate the energy of the one-dimensional infinite potential well is given as follows.

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\left(\frac{h^2}{8m{L}^2}\right){\mathit{n}}^{\mathbf{2}}$ …… (i)

Here,h is the Plank’s constant,m is the mass,n is the energy states and L is the width.

When the width of the potential well is double.

Substitute 2L for into L equation (i).

$$\begin{gathered} E_n^{\text{'}}=\left(\frac{h^2}{8m\times {\left(2L\right)}^2}\right)n^2 \\ {E}_n^{\text{'}}=\left(\frac{h^2}{8m\times 4L^2}\right)n^2 \\ {E}_n^{\text{'}}=\frac{1}{4}\left(\frac{h^2}{8m{L}^2}\right)n^2 \\ {E}_n^{\text{'}}=\frac{1}{4}{E}_n \end{gathered}$$

Hence the new energy is multiplied by the factor of 1/4.

(b)

Recall the equation (i),

$E_n=\left(\frac{h^2}{8mL}\right)n^2$

Form the above equation, it can be seen that the energy of the state is directly proportional to the energy states.

$E_n\alpha n^2$

Since the higher energy depends only on n.therefore, energies of higher energy states multiplied by the same factor.

Hence, the energies of higher energy states multiplied by the same factor.