An electron is trapped in a one-dimensional infinite potential well. For what (a) higher quantum number and (b) lower quantum number is the corresponding energy difference equal to the energy difference $\mathbf{∆}{\mathit{E}}_{\mathbf{43}}$ between the levels n = 4 and n = 3 ? (c) Show that no pair of adjacent levels has an energy difference equal to $\mathbf{2}\mathbf{∆}{\mathit{E}}_{\mathbf{43}}$ .

An electron is a negatively charged subatomic particle. It can be either free (not attached to any atom), or bound to the nucleus of an atom. Electrons in atoms exist in spherical shells of various radii, representing energy levels. The larger the spherical shell, the higher the energy contained in the electron.

Assume that the quantum numbers of the pairs be n and n + 1 .

The lowest quantum number is n that is, n = 3

And, the highest quantum umber is n + 1 . That is,

n + 1 = 4

Now, calculate the lowest quantum number as follow:

For the lowest quantum number, subtract the higher energy from the lowest as follows:

$$\begin{gathered} E_{n+1}-E_n=∆E_{43} \\ =E_4-E_3 \\ =\left(n_4^2-n_3^2\right)\left(\frac{{\pi }^2{h}^2}{2m\tau }\right) \\ =\left(4^2-3^2\right)\left(\frac{{\pi }^2{h}^2}{2m\tau }\right) \\ =7\left(\frac{{\pi }^2{h}^2}{2m\tau }\right) \end{gathered}$$

Also,

$$\begin{gathered} E_{n_1+n_2}=\left(n_1^2-n_2^2\right)\left(\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{\tau }^2}\right) \\ =∆E \\ =7\left(\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{\tau }^2}\right) \end{gathered}$$

Therefore,

$\left(n_1^2-n_2^2\right)=7$

Use the following formula to solve further,

$a^2-b^2=\left(a+b\right)\left(a-b\right)$

So, you can write,

$\begin{array}{l}\left(n_1^2-n_2^2\right)=7\\ (\left(n_1-n_2\right)\left(n_1+n_2\right)=7\end{array}$

Thus the different factors that can be made are as follow;

$$\begin{gathered} n_1-n_2=1 \\ {n}_1+n_2=7 \end{gathered}$$

Hence the highest quantum number is 7 .

The lower quantum number is,

$n_1-n_2=1$

Hence the lowest quantum number is 1 .

$n_1$To find $n_1$and $n_2$such that,

$$\begin{gathered} {}_{43}E_{n_1+n_2}=\left(\frac{h^2{\pi }^2}{2m{\tau }^2}\right)\left(n_1^2-n_2^2\right) \\ =2∆E_{43} \end{gathered}$$

Therefore,

$\frac{h^2{\pi }^2}{2m{\tau }^2}\left(n_1^2-n_2^2\right)=2∆E_{43}$ …… (1)

Now substitute 4 for $n_1$and 3 for $n_2$in the above equation.

$$\begin{gathered} \frac{h^2{\mathrm{\pi }}^2}{2m{\tau }^2}\left(4^2-3^2\right)=2∆E_{43} \\ \frac{h^2{\mathrm{\pi }}^2}{2m{\tau }^2}\left(16-9\right)=2∆E_{43} \\ \frac{7h^2{\mathrm{\pi }}^2}{2m{\tau }^2}=2∆E_{43} \end{gathered}$$

Therefore,

$$\begin{gathered} 2∆E_{43}=2\left(\frac{7h^2{\mathrm{\pi }}^2}{2m{\tau }^2}\right) \\ =\frac{14h^2{\mathrm{\pi }}^2}{2m{\tau }^2} \end{gathered}$$

Thus the equation $\frac{h^2{\pi }^2}{2m{\tau }^2}\left(n_1^2-n_2^2\right)=2∆E_{43}$

Implies,

$$\begin{gathered} \left(n_1^2-n_2^2\right)=14 \\ {a}^2-b^2=\left(a+b\right)\left(a-b\right),theequation \\ \left(n_1^2-n_2^2\right)=14isrewrittenasfollows: \\ \left(n_1-n_2\right)\left(n_1+n_2\right)=14 \end{gathered}$$

There are two ways in which 14 can be written one is $\left(2\times 7\right)$and the another is $\left(14\times 1\right)$.

Therefore,

$(n_1-n_2)\left(n_1+n_2\right)=\left[\begin{array}{c}\left(2\times 7\right)\\ (14\times 1)\end{array}\right]$

Thus, the different factor that can be made are as follows:

$n_1-n_2=2$ ….. (2)

$n_1+n_2=7$ ….. (3)

Solving the equation (2) and (3) for $n_1$as follows:

$2n_1=9$

$\begin{array}{l}{n}_1=\frac{9}{2}\\ =4.5\\ =notaninteger\end{array}$

Other factors are,

$n_1-n_2=1$ ….. (4)

$n_1+n_2=14$ ….. (5)

Solving the equation (5) and (5) for as follows:

$2n_1=15$

$\begin{array}{l}{n}_1=\frac{15}{2}\\ =7.5\\ =notaninteger\end{array}$

Hence, it is not possible that adjacent levels having an energy difference of $2∆E_{43}$.