### Parameters to Use for the Following Problems#### Fundamental Constants:- \( K = 1.38 \times 10^{-23} \, \text{J/K} \)- \( q = 1.6 \times 10^{-19} \, \text{coulombs} \)#### For Silicon at Room Temperature (\( T = 300^\circ K \)):- Energy Gap (\( E_G \)) = 1.12 eV- Intrinsic Carrier Concentration (\( n_i \)) = \( 1 \times 10^{10} \, \text{cm}^{-3} \)- Permittivity of Silicon (\( \epsilon_{Si} = K_{Si}\epsilon_0 \)) - \( K_{Si} = 11.8 \) - \( \epsilon_0 = 8.85 \times 10^{-14} \, \text{F/cm} \)#### Carrier Mobilities and Diffusion Constants:- Hole Mobility (\( \mu_p \)) = \( 384.62 \, \text{cm}^2/\text{V-sec} \)- Electron Mobility (\( \mu_n \)) = \( 1230.77 \, \text{cm}^2/\text{V-sec} \)- Hole Diffusion Constant (\( D_p \)) = \( 10 \, \text{cm}^2/\sec \)- Electron Diffusion Constant (\( D_n \)) = \( 32 \, \text{cm}^2/\sec \)#### Carrier Lifetimes:- Electron Lifetime (\( \tau_n \)) = \( 2.0 \times 10^{-4} \, \sec \)- Hole Lifetime (\( \tau_p \)) = \( 9.0 \times 10^{-5} \, \sec \)#### Carrier Diffusion Lengths:- Electron Diffusion Length (\( L_n \)) = \( 8.0 \times 10^{-2} \, \text{cm} \)- Hole Diffusion Length (\( L_p \)) = \( 3.0 \times 10^{-2} \, \text{cm} \)### Problem 1:A sample of silicon at room temperature and in thermal equilibrium is doped with donor atoms such that \( N_D = 4 \times

Solutions- Given data- Nd = 4 x 1015 cm-3 Ni = 1 x 1010 cm-3 T = 300K

Parameters to use for the problems that follow: K = 1.38 × 10-23 J/K q = 1.6 × 10-¹⁹ coulombs For Silicon at room temperature (T = 300°K): EG = 1.12eV -3 n₁ = 1 × 10¹0 cm- € Si Ksio (11.8) (8.85 × 10-¹4)F/cm = Hp 384.62cm²/V - sec, Dp = 10cm²/sec = 1230.77cm²/V – sec, - = fln Dn = 32cm²/sec Tn 1 2.0 × 10-¹sec, Tp = 9.0 × 10-5sec = 8.0 × 10-2cm, Ep = 3.0 × 10-²cm En - A sample of silicon at room temperature and in thermal equilibrium is doped with donor atoms such that ND = 4 x 10¹5 cm-³. Determine the concentration of electrons and holes. ND

Parameters to use for the problems that follow: K = 1.38 × 10-23 J/K q = 1.6 × 10-¹⁹ coulombs For Silicon at room temperature (T = 300°K): EG = 1.12eV -3 n₁ = 1 × 10¹0 cm- € Si Ksio (11.8) (8.85 × 10-¹4)F/cm = Hp 384.62cm²/V - sec, Dp = 10cm²/sec = 1230.77cm²/V – sec, - = fln Dn = 32cm²/sec Tn 1 2.0 × 10-¹sec, Tp = 9.0 × 10-5sec = 8.0 × 10-2cm, Ep = 3.0 × 10-²cm En - A sample of silicon at room temperature and in thermal equilibrium is doped with donor atoms such that ND = 4 x 10¹5 cm-³. Determine the concentration of electrons and holes. ND

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

4. b) A 0.15 M aqueous solution of some weak acid (HA) has a pH of 4.55 at 25 °C. What is the AG° for the ionization of this acid?
A Al(s) + 12(s) = All3(s) В CH4(g) + 4CI2(g) = C(4(1) + 4HCI(g) C Fe,O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g) Which of these processes is AH < 0 OB and C only OA and B only O A, B and OC only
n²a 3. a) In the van der Waals equation, (P +) (V – nb) = nRT, explain why a term is added to the measured pressure and why a term is subtracted from the container volume to V2 correct for non-ideal gas behaviour.
Question 9. Consider the following system with a potential function: V(x) = }. 0 (0 0, L 3L) E V(x)=0 (barrier) V(x)=0 (well A) V(x)=0 (barrier) V(x)=0 (well B) V(x)=0 (barrier) 2L 3L The diagram above is a sketch of this potential V(x). Compared to the "particle in a box" (PIB) system, this system can be called "particle in two boxes" (PITB) and its solution largely resembles that of the PIB. (a) Where would the wavefunction solutions exist (i.e., be non-zero)? Explain your reasoning. (b) How might we modify this potential energy diagram to become a "double-well" potential? Sketch such a potential. Where would the wavefunction solutions exist in your potential? (c) Now let us be more specific. Where would the wavefunction solutions exist for small quantum numbers, n? What happens in the range L
Construct enthalpy cycles; use Hess's law and the enthalpy changes of formation below to calculate the enthalpy change for the reaction, CaCos(s) – ÇaQ(s) + CO2(g) AH, Cacoa(s) = -1207 kJmol-1 AH Cao(s) = -635 kJmol AH CO2(9) = -393 kJmol 1