Question

Draw an energy-level diagram, similar to Figure 38.21, for the ${\mathrm{He}}^{+}$ion. On your diagram:

a. Show the first five energy levels. Label each with the values of n and${\mathrm{E}}_{\mathrm{n}}$

b. Show the ionization limit.

c. Show all possible emission transitions from the n = 4 energy level.

d. Calculate the wavelengths (in nm) for each of the transitions in part c and show them alongside the appropriate arrow.

Short answer

(a) The set distances between electrons and the nucleus of an atom are known as energy levels.

(b) The ionisation limit is represented by$\mathrm{n}=\mathrm{\infty }$

(c) On the diagram, all conceivable emission transitions from the n=4 energy level are depicted.

(d) The wavelengths are${\mathrm{\lambda }}_{43}=477\mathrm{nm};{\mathrm{\lambda }}_{42}=122\mathrm{nm};{\mathrm{\lambda }}_{41}=24\mathrm{nm}$${\mathrm{\lambda }}_{43}=477\mathrm{nm};{\mathrm{\lambda }}_{42}=122\mathrm{nm};{\mathrm{\lambda }}_{41}=24\mathrm{nm}$

Step-by-step solution

Given information

Consider the values are given in n and${\mathrm{E}}_{\mathrm{n}}.$

Part(a) Step 2: Creating Energy level diagram and calculations

We can begin by looking at expression energy levels:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=-{\mathrm{Z}}^2\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2} \\ \mathrm{Z}=2\left(\text{for}{\mathrm{He}}^{+}\right) \\ {\mathrm{E}}_{\mathrm{n}}=-2^2\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\text{(substitute}\mathrm{Z}=2\text{)} \\ {\mathrm{E}}_{\mathrm{n}}=-\frac{54.4\mathrm{eV}}{{\mathrm{n}}^2} \end{gathered}$$

We get the following energies by substituting n from 1 to 5: -54.4 eV,-13.6 eV,-6.0 eV,-3.4 eV,-2.2 eV are the values shown in the diagram.

Part (b) Step 1: Given information

From alkali metals to noble gases, the ionisation energy gradually increases. Due to the increasing distance of the valence electron shell from the nucleus, the maximum ionisation energy falls from the first to the final row in a particular column. For items with a value greater than 104, predicted values are utilised.

Part (b) Step 2: 

The ionisation limit is represented by $\mathrm{n}=\infty$

Part (c) Step 1: 

Consider the emission transitions energy level is n = 4

Part (c) Step 2: Representing Energy level n=∞

On the diagram, all conceivable emission transitions from the n=4 energy level are illustrated, and they include:$4\to 3,4\to 2,4\to 1$

Part (d) Step 1: Given information

Planck's equation connects the energy change associated with a transition to the electromagnetic wave's frequency is${\mathrm{\lambda }}_{\mathrm{nm}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_{\mathrm{m}}}$

Part (d) Step 2: Calculations

The following expression can now be used:

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{nm}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_{\mathrm{m}}} \\ {\mathrm{\lambda }}_{43}=\frac{\mathrm{hc}}{{\mathrm{E}}_4-{\mathrm{E}}_3}(\text{emission}4\to 3) \\ {\mathrm{\lambda }}_{43}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{-3.4\cdot 1.6\cdot {10}^{-19}-\left(-6.0\cdot 1.6\cdot {10}^{-19}\right)}\text{(substitute)} \\ {\mathrm{\lambda }}_{43}=477\mathrm{nm} \\ {\mathrm{\lambda }}_{42}=\frac{\mathrm{hc}}{{\mathrm{E}}_4-{\mathrm{E}}_2}(\text{emission}4\to 2\text{)} \\ {\mathrm{\lambda }}_{42}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{-3.4\cdot 1.6\cdot {10}^{-19}-\left(-13.6\cdot 1.6\cdot {10}^{-19}\right)}\text{(substitute)} \\ {\mathrm{\lambda }}_{42}=122\mathrm{nm} \\ {\mathrm{\lambda }}_{41}=\frac{\mathrm{hc}}{{\mathrm{E}}_4-{\mathrm{E}}_1}(\text{emission}4\to 1) \\ {\mathrm{\lambda }}_{41}=\frac{6.63\cdot {10}^{-34}\cdot 3\cdot {10}^8}{-3.4\cdot 1.6\cdot {10}^{-19}-\left(-54.4\cdot 1.6\cdot {10}^{-19}\right)}\text{(substitute)} \\ {\mathrm{\lambda }}_{41}=24\mathrm{nm} \end{gathered}$$