Chapter 28, Problem 48GP

(a)

To determine

The effective charge ${\text{Z}}_{\text{eff}}$ seen by this electron.

Answer to Problem 48GP

The answer is ${\text{Z}}_{\text{eff}}=1.56$

Explanation of Solution

Given info: Binding energy of outermost electrons in Boron is $8.26eV$

Formula used:

Energy levels of $\text{H}$ atom are given by ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}$

Where, $\text{n}$ is the principal quantum number.

Calculation:

Boron has $\text{Z = 4}$ . So, the outermost electron has $\text{n=2}$

Use the Bohr result with an effective $\text{Z}$ :

  $\begin{array}{l}{\text{E}}_{\text{2}}\text{= -}\frac{\left(\text{13}\text{.6}\text{eV}\right){\left({\text{Z}}_{\text{eff}}\right)}^{\text{2}}}{{\text{n}}^{\text{2}}}\\ \text{-8}\text{.26}\text{eV}\text{=}\text{ -}\frac{\left(\text{13}\text{.6}\text{eV}\right){\left({\text{Z}}_{\text{eff}}\right)}^{\text{2}}}{{\text{2}}^{\text{2}}}\\ {\text{Z}}_{\text{eff}}\text{=}\text{1}\text{.56}\end{array}$

Understand that it indicates some shielding by second electron in the $\text{n = 2}$ shell.

(b)

To determine

The average orbital radius.

Answer to Problem 48GP

The answer is $\text{1}\text{.4}\text{×}{\text{10}}^{\text{-10}}\text{m}$

Explanation of Solution

Given info: Electron on boron is $8.26eV$

Effective charge, ${\text{Z}}_{\text{eff}}=1.56$

Formula used:

It is known that $\text{r = }\frac{{\text{n}}^{\text{2}}{\text{r}}_{\text{1}}}{{\text{Z}}_{\text{eff}}}$

Where, $\text{n}$ is the principal quantum number, ${\text{r}}_{\text{1}}$ is a Bohr radius and ${\text{Z}}_{\text{eff}}$ is the effective charge

Calculation:

Boron has $\text{Z = 4}$ . So, the outermost electron has $\text{n=2}$

Substitute all the given values in expression $\text{r = }\frac{{\text{n}}^{\text{2}}{\text{r}}_{\text{1}}}{{\text{Z}}_{\text{eff}}}$

  $\text{r = }\frac{{\text{2}}^{\text{2}}\left(\text{0}\text{.529}{\text{×10}}^{\text{-10}}\text{m}\right)}{\text{1}\text{.56}}\text{=1}\text{.4}\text{×}{\text{10}}^{\text{-10}}\text{m}$