91Ó°ÊÓ

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

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.

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.

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

Consider the emission transitions energy level is n = 4

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

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

The following expression can now be used:

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