Question

As a crude approximation, an impurity pentavalent atom in a (tetravalent) silicon lattice can be treated as a one-electron atom, in which the extra electron orbits a net positive charge of 1. Because this "atom" is not in free space, however, the permitivity of free space, ${\mathit{\epsilon }}_{\mathbf{0}}$. must be replaced by $\mathit{\kappa }{\mathit{\epsilon }}_{\mathbf{0}}$, where$\mathit{\kappa }$ is the dielectric constant of the surrounding material. The hydrogen atom ground-state energies would thus become

$\mathit{E}\mathbf{=}\mathbf{-}\frac{\mathbf{m}{\mathbf{e}}^{\mathbf{4}}}{\mathbf{2}{\mathbf{\left(}\mathbf{4}\mathbf{\pi }\mathbf{\kappa }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\right)}}^{\mathbf{2}}{\mathbf{\hslash }}^{\mathbf{2}}}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{-}\mathbf{13}\mathbf{.6}\mathbf{\text{eV}}}{{\mathbf{\kappa }}^{\mathbf{2}}{\mathbf{n}}^{\mathbf{2}}}$

Given $\mathit{\kappa }\mathbf{=}\mathbf{12}$for silicon, how much energy is needed to free a donor electron in its ground state? (Actually. the effective mass of the donor electron is less than , so this prediction is somewhat high.)

Short answer

The amount of energy is needed to free a donor electron in its ground state is$-0.094\text{eV}$.

Step-by-step solution

Given data

$k=12$- dielectric constant

The energy of hydrogen atom in$n$state is,$E=\frac{-13.6\text{eV}}{k^2{n}^2}$.

Formula of energy of hydrogen

^{The energy of hydrogen atom in $\mathit{n}$ state is, $\mathit{E}\mathbf{=}\frac{\mathbf{-}\mathbf{13}\mathbf{.6}\mathbf{\text{eV}}}{{\mathbf{k}}^{\mathbf{2}}{\mathbf{n}}^{\mathbf{2}}}$.}

Step 3:Determine the amount of energy required to free a donor electron 

The dielectric constant for silicon atom is,$k=12$.

Substitute 12 for$k$and 1 for$n$in equation$E=\frac{-13.6eV}{k^2{n}^2}$.

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

The amount of energy is needed to free a donor electron in its ground state is $-0.094\text{eV}$.