Question

A hydrogenic atom consists of a single electron orbiting a nucleus with Z protons. ($\mathbf{Z}\mathbf{=}\mathbf{1}$ would be hydrogen itself,$\mathit{Z}\mathbf{=}\mathbf{2}$is ionized helium ,$\mathit{Z}\mathbf{=}\mathbf{3}$is doubly ionized lithium, and so on.) Determine the Bohr energies $\mathit{E}\mathit{n}\mathbf{\left(}\mathit{Z}\mathbf{\right)}$, the binding energy${\mathit{E}}_{\mathbf{1}}\mathbf{\left(}\mathit{Z}\mathbf{\right)}$, the Bohr radius$\mathit{a}\mathbf{\left(}\mathit{Z}\mathbf{\right)}$, and the Rydberg constant R$\mathbf{\left(}\mathit{Z}\mathbf{\right)}$for a hydrogenic atom. (Express your answers as appropriate multiples of the hydrogen values.) Where in the electromagnetic spectrum would the Lyman series fall, for $\mathit{Z}\mathbf{=}\mathbf{2}$and $\mathit{Z}\mathbf{=}\mathbf{3}$? Hint: There’s nothing much to calculate here— in the potential (Equation 4.52) $\mathit{Z}{\mathit{e}}^{\mathbf{2}}$, so all you have to do is make the same substitution in all the final results.

$\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }\dot{{\mathbf{o}}_{\mathbf{0}}}}\frac{\mathbf{1}}{\mathbf{r}}$ (4.52).

Short answer

The electronic spectrum would be the Lyman seires fall, for$z=2$ and$z=3$ is at,$z=2$ is$3.04\times {10}^{-8}m.$

At $z=3$is$1.35\times {10}^{-8}m.$

Step-by-step solution

Given

The potential is given by:

$V\left(r\right)=-\frac{e^2}{4\mathrm{\pi }\dot{{\mathrm{o}}_0}}\frac{1}{r}$

Electronic spectrum

Consider a hydrogen, if irradiated with light it will be excited to another energy level. The energies of these photons are easily calculated from the Bohr energy formula:

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}\frac{\mathbf{m}{\mathbf{e}}^{\mathbf{4}}}{\mathbf{2}{\mathbf{h}}^{\mathbf{2}}{\mathbf{(}\mathbf{4}\mathbf{{\rm T}}\mathbf{{\rm T}}\mathbf{}{\mathbf{\in }}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}}$

Lyman lines range from${\mathit{n}}_{\mathbf{i}}\mathbf{=}\mathbf{2}\mathbf{}\mathit{t}\mathit{o}\mathbf{}{\mathit{n}}_{\mathbf{i}}\mathbf{=}\mathbf{\infty }\mathbf{(}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathbf{}{\mathit{n}}_{\mathbf{f}}\mathbf{=}\mathbf{1}\mathbf{)}$;

the wavelengths range from

Finding the electronic spectrum

$$\begin{gathered} \frac{1}{{\lambda }_2}=R\left(1-\frac{1}{4}\right) \\ =\frac{3}{4}R \\ {\lambda }_2=\frac{4}{3R} \end{gathered}$$

Down to

role="math" localid="1658210586427" $$\begin{gathered} \frac{1}{{\lambda }_1}=R\left(1-\frac{1}{\infty }\right) \\ =R \\ \Rightarrow R_1=\frac{1}{R} \end{gathered}$$

For$Z=2$

$$\begin{gathered} {\lambda }_1=\frac{1}{4R} \\ =\frac{1}{4(1.097\times {10}^7)} \\ =2.28\times {10}^{-8}m \end{gathered}$$

to

$$\begin{gathered} {\lambda }_2=\frac{1}{3R} \\ =3.04\times {10}^{-8}m, \end{gathered}$$

Ultraviolet.