91Ó°ÊÓ

The given data is listed below as-

The formula for finding the energy of photon when a hydrogen atom jumps from a state with principal quantum number 2 to a state with principal quantum number 1 when${\mathit{n}}_{\mathbf{1}}\mathbf{<}{\mathit{n}}_{\mathbf{2}}$is given by-

$\mathit{E}\mathbf{=}\mathit{A}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Here, A = 13.6 ev.

The energy absorbed by the atom is given by-

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=1$ and $n_2=4$

$\begin{array}{l}E=\left(13.6ev\right)\left(\frac{1}{1^2}-\frac{1}{4^2}\right)\\ =12.75ev\\ E=12.75ev\end{array}$

Thus, the energy absorbed by the atom is $12.75ev$.

When the atom is de-excited to the ground state, there are six possible ways to do this. The transitions are:

$4→34→24→13→23→12→1$

Thus, there are six possible ways.

The highest energy difference occurs at $4→1$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=1$ and $n_2=4$

$$\begin{gathered} E_{4→1}=(13.6ev)\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ =12.75ev \\ {E}_{4→1}=12.75ev \end{gathered}$$

Thus, the highest energy is 12.75 ev.

The second highest energy difference occurs at $3→1$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=1$ and $n_2=3$

$$\begin{gathered} E_{4→1}=(13.6ev)\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ =12.1ev \\ {E}_{3→1}=12.1ev \end{gathered}$$

Thus, the second highest energy is 12.1 ev.

The third highest energy difference occurs at $2→1$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=1$ and $n_2=2$

$$\begin{gathered} E=(13.6ev)\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ =10.2ev \\ {E}_{2→1}=10.2ev \end{gathered}$$

Thus, the third highest energy is 10.2 ev .

The lowest energy difference occurs at $4→3$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=3$ and $n_2=4$

$$\begin{gathered} E=(13.6ev)\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ =0.661ev \\ {E}_{4→3}=0.661ev \end{gathered}$$

Thus, the lowest energy is 0.661 ev .

The second lowest energy difference occurs at $3→2$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=2$ and $n_2=3$

$$\begin{gathered} E=(13.6ev)\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ =1.89ev \\ {E}_{3→2}=1.89ev \end{gathered}$$

Thus, the second lowest energy is 1.89 ev .

The third lowest energy difference occurs at $4→2$

$E=A\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For, $n_1=2$ and $n_2=4$

$$\begin{gathered} E=(13.6ev)\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ =2.55ev \\ {E}_{4→2}=2.55ev \end{gathered}$$

Thus, the third lowest energy is 2.55 ev.