91Ó°ÊÓ

Consider the exponential formula as per the Boltzmann’s distribution:

${\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{E}}_{\mathrm{gap}}}{\mathbf{2}{\mathbf{k}}_b\mathbf{T}}}$ ….. (1)

Here,$k_b$ is Boltzmann constant and the temperature is T.

Consider the expression for the vibrational distance as:

Substitute the values in the equation () and solve as:

$\begin{array}{c}{e}^{\frac{−E_{gap}}{2k_{b}T}}=e^{\frac{−1\text{ e³Õ}}{2×1.38×{10}^{−23}×300}}\text{ e³Õ}\\ =\text{4}×{10}^{−9}\text{ e³Õ}\end{array}$

Note the availability of the electron is less than $\text{4}×{10}^{−9}\text{ e³Õ}$so the resistivity is approximately$\frac{1}{\text{4}×{10}^{−9}\text{ e³Õ}}=2.5×{10}^8$

Consider the equation for the ration between the resistivity of the silicon and the silver as:

$\frac{{\mathrm{ÒÏ}}_{silicon}}{{\mathrm{ÒÏ}}_{silver}}$

Substitute the values and solve as:

$\begin{array}{c}\frac{{\mathrm{ÒÏ}}_{silicon}}{{\mathrm{ÒÏ}}_{silver}}=\frac{1.6×{10}^{−8}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}}{10\text{ }\mathrm{Î\copyright }â‹\dots \text{m}}\\ =1.6×{10}^{−9}\end{array}$

Therefore, the factor is approximately similar to the ratio between the resistivity.

Since, the energy gap of the diamond is$−5\text{ e³Õ}$ .

Solve for the gap as:

$\begin{array}{c}{E}_{gap}=−5\text{ e³Õ}\\ =(−5\text{ e³Õ})\left(\frac{1.6×{10}^{−19}\text{ J}}{1\text{ e³Õ}}\right)\\ =(−5)(1.6×{10}^{−19}\text{ J})\end{array}$

Consider the formula for the resistivity of the diamond as:

${\mathrm{ÒÏ}}_{diamond}=e^{\frac{−E_{gap2}}{2k_{b}T}}\frac{{\mathrm{ÒÏ}}_{silicon}}{{\mathrm{ÒÏ}}_{silver}}$

Resolve the equation as:

${\mathrm{ÒÏ}}_{diamond}=\frac{{\mathrm{ÒÏ}}_{silver}}{e^{\frac{−E_{gap2}}{2k_{b}T}}}$

Substitute the values and solve as:

$\begin{array}{c}{\mathrm{ÒÏ}}_{diamond}=\frac{1.6×{10}^{−8}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}}{e^{\frac{−\left(5\right)(1.6×{10}^{−19}\text{ J})}{2×1.38×{10}^{−23}×300}}}\\ =\frac{1.6×{10}^{−8}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}}{{10}^{−42}}\\ =1.6×{10}^{−8}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}\left({10}^{42}\right)\\ =1.6×{10}^{34}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}\end{array}$

Therefore, the resistivity of the diamond is$1.6×{10}^{34}\text{ }\mathrm{Î\copyright }â‹\dots \text{m}$ .