Chapter 29, Problem 39GP

(a)

To determine

The binding energy.

Answer to Problem 39GP

The binding energy is $E_n=-0.094\text{ eV}$

Explanation of Solution

Given:

Dielectric constant, $K=12$

Formula used:

Binding energy of an electron is given by:

  $E_n=\left(-13.6\frac{Z^2}{K^2{n}^2}\right)\text{eV}$

Calculation:

For $n=1$ , $Z=1$

Hence, the binding energy is given by:

  $\begin{array}{l}{E}_n=\left(-13.6\frac{{\left(1\right)}^2}{{\left(12\right)}^2{\left(1\right)}^2}\right)\text{eV}\\ E_n=-0.094\text{ eV}\end{array}$

Conclusion:

Thus, the binding energy is $E_n=-0.094\text{ eV}$

(b)

To determine

Orbit radius for the extra electron.

Answer to Problem 39GP

Orbital radius is $r_n=0.64\text{ nm}$

Explanation of Solution

Given:

Dielectric constant, $K=12$

Formula used:

Orbital radius is given by:

  $r_n=\frac{K{n}^2{\hslash }^2}{Z{e}^{2}mk}$

Calculation:

Substituting all the values, in the above formula,

  $\begin{array}{l}{r}_n=\frac{K{n}^2{\hslash }^2}{Z{e}^{2}mk}\\ r_n=\frac{K{n}^2}{Z}\left(0.53\stackrel{°}{\text{A}}\right)\\ r_n=\frac{\left(12\right){\left(1\right)}^2}{1}\left(0.53\times {10}^{-10}\text{m}\right)\\ r_n=0.64\text{ nm}\end{array}$

Conclusion:

Thus, the orbital radius is $r_n=0.64\text{ nm}$