Your answer Activate Window File unlnadinn will he availahle after the comnletion of the testina of the second stane please enter at lanet 1 character in the tert field and save the answer Silicon (Si) was doped by arsenic (As) with concentration of 10^16at/cm^3 Estimate the conductivity of the silicon at the temperature 450 K, if mobility of electrons (l_(e)) at given temperature is 2000cm^2/(V^ast s) mobility of holes (I_(p)) is 300cm^2/(V^ast s) ,conductivity of pure Si at 300 K is 5^ast 10^-50hm^-1cm^-1 , and bandgap (Eg) for Si is 1.1 eV.Give the answer in Ohm^-1cm^-1 with rounding to 2 significant digits. Be careful:the assessment will take into account the progress of the solution;writing only the answer is not enough.

To estimate the conductivity of the silicon at 450 K, we need to consider the contributions of both electrons and holes to the overall conductivity. The conductivity due to electrons and holes can be calculated using the formula:<br /><br />\[ \sigma = \sigma_e + \sigma_h \]<br /><br />where:<br />- \( \sigma_e \) is the conductivity due to electrons<br />- \( \sigma_h \) is the conductivity due to holes<br /><br />The conductivity due to electrons and holes can be calculated using the following formulas:<br /><br />\[ \sigma_e = n_e \cdot e \cdot \mu_e \]<br />\[ \sigma_h = n_h \cdot e \cdot \mu_h \]<br /><br />where:<br />- \( n_e \) is the electron concentration<br />- \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \) C)<br />- \( \mu_e \) is the mobility of electrons<br />- \( n_h \) is the hole concentration<br />- \( \mu_h \) is the mobility of holes<br /><br />Given that the silicon is doped with arsenic, we can assume that the majority carriers are electrons and the minority carriers are holes. Therefore, we can use the following relationships:<br /><br />\[ n_e = N_D - N_A \]<br />\[ n_h = N_A \]<br /><br />where:<br />- \( N_D \) is the doping concentration of electrons<br />- \( N_A \) is the doping concentration of holes<br /><br />Given that the doping concentration is \( 10^{16} \, \text{at/cm}^3 \), we can assume that the majority carriers are electrons and the minority carriers are holes. Therefore, we can use the following relationships:<br /><br />\[ n_e = 10^{16} \, \text{cm}^{-3} \]<br />\[ n_h = 10^{16} \, \text{cm}^{-3} \]<br /><br />Now, we can calculate the conductivity due to electrons and holes:<br /><br />\[ \sigma_e = n_e \cdot e \cdot \mu_e \]<br />\[ \sigma_e = 10^{16} \, \text{cm}^{-3} \cdot 1.6 \times 10^{-19} \, \text{C} \cdot 2000 \, \text{cm}^2/\text{V} \cdot \text{s} \]<br />\[ \sigma_e = 3.2 \times 10^{-3} \, \text{S/cm} \]<br /><br />\[ \sigma_h = n_h \cdot e \cdot \mu_h \]<br />\[ \sigma_h = 10^{16} \, \text{cm}^{-3} \cdot 1.6 \times 10^{-19} \, \text{C} \cdot 300 \, \text{cm}^2/\text{V} \cdot \text{s} \]<br />\[ \sigma_h = 4.8 \times 10^{-3} \, \text{S/cm} \]<br /><br />Finally, we can calculate the total conductivity:<br /><br />\[ \sigma = \sigma_e + \sigma_h \]<br />\[ \sigma = 3.2 \times 10^{-3} \, \text{S/cm} + 4.8 \times 10^{-3} \, \text{S/cm} \]<br />\[ \sigma = 8.0 \times 10^{-3} \, \text{S/cm} \]<br /><br />Therefore, the estimated conductivity of the silicon at 450 K is \( 8.0 \times 10^{-3} \, \text{S/cm} \).