Chapter 36, Problem 65P

To determine

The transition that results in the spectral lines.

Answer to Problem 65P

The transition from $3\to 2$ gives wavelength of $164\text{nm}$ , the transition from $9\to 3$ gives wavelength of $230.06\text{nm}$ and the transition from $7\to 4$ gives the wavelength of $541\text{nm}$ .

Explanation of Solution

Given data:

The given values of the wavelength $\lambda$ are $164\text{nm}$ , $230.06\text{nm}$ and $541\text{nm}$ .

Formula:

The expression to determine the energy $E_{\text{n}}$ of the electron in the $n^{\text{th}}$ state of the atom is given by,

  $E_{\text{n}}=-Z^2\frac{E_0}{n^2}$

The expression to determine the wavelength of the $\lambda$ of a spectral line associated with the transition between two energy states of an atom is given by,

  $\lambda =\frac{1240\text{eV}\cdot \text{Nm}}{\Delta E}$

The expression for the energy difference $\Delta E$ between the two states is given by,

  $\begin{array}{c}\Delta E=E_{\text{i}}-E_{\text{f}}\\ =-Z^2{E}_0\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\end{array}$

Calculation:

The expression for the energy difference $\Delta E$ between the two states by keeping the atomic number of helium as $Z$ equal to $2$ is evaluated as,

  $\begin{array}{c}\Delta E=Z^2{E}_0\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\ ={\left(2\right)}^2\left(13.6\text{eV}\right)\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\ =54.4\text{eV}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\end{array}$

The wavelength of the $\lambda$ of a spectral line associated with the transition between two energy states of an atom is calculated as,

  $\begin{array}{l}\lambda =\frac{1240\text{eV}\cdot \text{Nm}}{\Delta E}\\ \frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{1240\text{eV}\cdot \text{Nm}}{54.4\text{eV}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)}\\ \frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{22.8\text{nm}}{164\text{nm}}\\ \frac{1}{n_{\text{f}}^2}=\frac{1}{n_{\text{i}}^2}+0.139\end{array}$

Solve further as,

  $n_{\text{f}}=\sqrt{\frac{n_{\text{i}}^2}{1+0.139n_{\text{i}}^2}}$

From above the table for the values of the final energy state for different values of the initial energy state is shown in Table 1

The required table is shown in Table 1

Table 1

$n_{\text{i}}$	$n_{\text{f}}$
$0$	$0$
$1$	$0.936887$
$2$	$1.603338$
$3$	$2$
$4$	$2.22773$
$5$	$2.363597$
$6$	$2.448674$
$7$	$2.543081$
$8$	$2.570484$
$9$	$2.570484$
$10$	$2.590639$

From above table it is clear that the transition is produced from state $3\to 2$ .

The state of transition when the for wavelength of $230.06\text{nm}$ is evaluated as,

  $\begin{array}{l}\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{22.8\text{nm}}{\lambda }\\ \frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{22.8\text{nm}}{230.06\text{nm}}\\ \frac{1}{n_{\text{f}}^2}=\frac{1}{n_{\text{i}}^2}+0.098872\\ n_{\text{f}}=\sqrt{\frac{n_{\text{i}}^2}{1+0.098872n_{\text{i}}^2}}\end{array}$

From above the table for the values of the final energy state for different values of the initial energy state is shown in Table 2

The required table is shown in Table 2

Table 2

$n_{\text{i}}$	$n_{\text{f}}$
$0$	$0$
$1$	$0.953952$
$2$	$1.693038$
$3$	$2.182264$
$4$	$2.489346$
$5$	$2.68344$
$6$	$2.809939$
$7$	$2.895443$
$8$	$2.955303$
$9$	$3$
$10$	$3.030686$

From above table it is clear that the transition is produced from state $9\to 3$ .

The state of transition when the for wavelength of $541\text{nm}$ is evaluated as,

  $\begin{array}{l}\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{22.8\text{nm}}{\lambda }\\ \frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}=\frac{22.8\text{nm}}{541\text{nm}}\\ \frac{1}{n_{\text{f}}^2}=\frac{1}{n_{\text{i}}^2}+0.04241\\ n_{\text{f}}=\sqrt{\frac{n_{\text{i}}^2}{1+0.04241n_{\text{i}}^2}}\end{array}$

From above the table for the values of the final energy state for different values of the initial energy state is shown in Table 3

The required table is shown in Table 3

Table 3

$n_{\text{i}}$	$n_{\text{f}}$
$0$	$0$
$1$	$0.979571$
$2$	$1.850126$
$3$	$2.55442$
$4$	$3.091309$
$5$	$3.489085$
$6$	$3.781754$
$7$	$4$
$8$	$4.160561$
$9$	$4.283982$
$10$	$4.379223$

From above table it is clear that the transition is produced from state $7\to 4$ .

Conclusion:

Therefore, the transition from $3\to 2$ gives wavelength of $164\text{nm}$ , the transition from $9\to 3$ gives wavelength of $230.06\text{nm}$ and the transition from $7\to 4$ gives the wavelength of $541\text{nm}$ .