91Ó°ÊÓ

We have been given silicon doped with phosphorus atoms.

Chemical potential is given by$\mathrm{μ}=-kT\ln \left(\frac{V{Z}_{int}}{N_c{v}_Q}\right)$

Due to two spin states $\mathrm{μ}=-kT\ln \left(\frac{2V}{N_c{v}_Q}\right)$

Ratio of Conduction electrons to number of doner atoms is $\frac{N_c}{N_d}=x=1-2e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)$

where $\mathrm{μ}=-kT\ln \left(\frac{2V}{x{N}_d{v}_Q}\right)$and $x=1-2e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)$

Let $t=\frac{kT}{I}$

Therefore Quantum volume $v_Q={\left(\frac{h^2}{2\mathrm{Ï€}mI}\right)}^{3/2}\frac{1}{t^{3/2}}$

Putting the vlaues of universal constants $m,handI.$$v_Q={\left(\frac{{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{2\mathrm{π}\left(9.11×{10}^{-31}\mathrm{kg}\right)\left(0.044×1.6×{10}^{-19}\mathrm{J}\right)}\right)}^{3/2}\frac{1}{t^{3/2}}$$=\left(3.6×{10}^{-26}\right)t^{-3/2}$

Also it is known $\frac{N_d}{V}=1×{10}^{23}{\mathrm{m}}^3$

Evaluting$\frac{\mathrm{μ}}{I}=-t\ln \left(\frac{2}{x\left(1×{10}^{23}{\mathrm{m}}^3\right)\left(3.6×{10}^{-26}\right)t^{-3/2}}\right)$

Therefore$\frac{\mathrm{μ}}{I}=-t\ln \left(\frac{2t^{3/2}}{0.0036x}\right)andx=1-\frac{(0.0036)e^{1/t}}{t^{3/2}}$.

Plotting a graph between $\raisebox{1ex}{$\mathrm{μ}$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$

We have to find whether the conduction elements can be treated as an ordinary ideal gas.

Considering the denominator of Fermi Dirac that is much greater than 1 we treat it as gas.

Therefore energies of the conduction band $\mathrm{Ï\mu }-\mathrm{μ}≫kT$

The difference between edge energy of conduction band is 3, therefore localid="1650464758608" $\mathrm{Ï\mu }-\mathrm{μ}≈3I$.

Solving $\frac{I}{kT}=\frac{0.044\mathrm{eV}}{0.026\mathrm{eV}}=1.695$

$\frac{\mathrm{Ï\mu }-\mathrm{μ}}{kT}≈\frac{3I}{kT}=3×1.695≈5$

Therefore the value of exponential $e^5≈150$which is much greater than 1.

Therefore approximating the Fermi Dirac distribution by Boltzmann distribution should be accurate.

We have to find out the temperature for which number of valence electrons excited to the conduction band would become comparable to the number of conduction electrons from donor impurities.

Saturation point of number of conduction electrons${10}^{17}\mathrm{P}\text{atoms per}{\mathrm{cm}}^3$

i.e$\frac{N_d}{V}=1×{10}^{23}{\mathrm{m}}^3$

For temperature $\frac{N_C}{V}=\frac{2}{v_Q}{e}^{-\mathrm{Δ}\mathrm{Ï\mu }/2kT}$

Let this expression be equal to ${10}^{23}$; Solving$-\frac{\mathrm{Δ}\mathrm{Ï\mu }}{2kT}=\ln \left(\frac{v_Q{N}_C}{2V}\right)$

$\frac{1}{T}=-\frac{2k}{\mathrm{Δ}\mathrm{Ï\mu }}\ln \left(\frac{v_Q{N}_C}{2V}\right)$

Substituting $\frac{1}{T}=-\frac{2\left(8.62×{10}^{-5}\mathrm{eV}\right)}{1.11\mathrm{eV}}\ln \left(\frac{v_Q{N}_C}{2V}\right)$

Quantum volume is given by $v_Q={\left(\frac{{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{2\mathrm{π}\left(9.11×{10}^{-31}\mathrm{kg}\right)\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)}\right)}^{3/2}\frac{1}{T^{3/2}}$

$=\left(3.6×{10}^{-20}\right)T^{-3/2}$

Therefore

$\frac{1}{T}=-\left(1.553×{10}^{-4}\right)\ln \left((0.0018)T^{-3/2}\right)$

Intersection in graph occurs at $T=520K$