Question

Starting from Equation 38.32, derive Equation 38.33

Short answer

The Derive equation is${\mathrm{E}}_{\mathrm{n}}=-\frac{1}{{\mathrm{n}}^2}\left(\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{2{\mathrm{a}}_{\mathrm{B}}}\right)$

Step-by-step solution

Given information 

In atomic physics, the Bohr radius is a physical constant. It discusses the most likely distance between a hydrogen atom's nucleus and electron in its ground state.

Calculations

We can begin with the expression (38.32) and the Bohr radius expression.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\frac{1}{2}\mathrm{m}\left(\frac{{\mathrm{ħ}}^2}{{\mathrm{m}}^2{\mathrm{a}}_{\mathrm{B}}^2{\mathrm{n}}^2}\right)-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0{\mathrm{n}}^2{\mathrm{a}}_{\mathrm{B}}} \\ {\mathrm{a}}_{\mathrm{B}}=\frac{4{\mathrm{\pi ϵ}}_0{\mathrm{ħ}}^2}{{\mathrm{me}}^2} \\ \frac{{\mathrm{ħ}}^2}{{\mathrm{ma}}_{\mathrm{B}}}=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0}\left(\mathrm{express}\right) \\ {\mathrm{E}}_{\mathrm{n}}=\frac{1}{2}\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0{\mathrm{n}}^2{\mathrm{a}}_{\mathrm{B}}}-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0{\mathrm{n}}^2{\mathrm{a}}_{\mathrm{B}}}\text{(substitute previous expression)} \\ {\mathrm{E}}_{\mathrm{n}}=-\frac{1}{{\mathrm{n}}^2}\left(\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{2{\mathrm{a}}_{\mathrm{B}}}\right) \end{gathered}$$