### 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

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 9RQ: What characterizes an electrolytic cell? What is an ampere? When the current applied to an...

See similar textbooks

Similar questions

Kk.73.
keCovalentActivity.do?locator assignment-take dr.Forms | Office of t... KB Viewing Your Aca... Have Changes in... Scholarship Ameri... bio 1108 chat [Review Topics] [References) Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH = -111 kJ, and K = 0.159, at 7.23 x 102 K: N₂ (9) + 3H₂(g) → 2NH3(g) req If the temperature on the equilibrium system is suddenly decreased: The value of K # 3 Oincreases Odecreases O remains the same E D C The value of Q O is less than Ke Ois greater than Ke O is equal to Ke The reaction must Orun in the forward direction to reestablish equilibrium Orun in the reverse direction to reestablish equilibrium O remain in the current position, since it is already at equilibrium The concentration of H₂ will O increase Odecrease Oremain the same 20 80, Submit Ans Show Hint ry Entire Group $ 888 FA F4 R F V % 5 T Cengage Learning | Cengage Technical Support FS B ^ 6 MacBook Air Y 9 more group…
doll all , if not then dot do skip
3 eq req For the reaction N₂ (9)+ 3H2(g) → 2NH3 (9) AH = -92.2 kJ and AS = -199 J/K The equilibrium constant for this reaction at 256.0 K is Assume that AH and AS are independent of temperature. Submit Answer $ R References to access important values if needed for this question. Retry Entire Group 9 more group attempts remaining Show Hint Cengage Learning | Cengage Technical Support 7 Z U * 8 1 M ( K O L Previous Next> Save an 5
d. For the reaction 2NO(g) → N2(9) + 02 (g) Species S J/(mol K) NO(g) 210.6 N2{9) 191.6 02(g) 205.0 AS = JK
A reaction Occuring in a Star 2 b 3 c H + H---> He + n a 1 2 d In order to balance the equation above, the values of a,b,c,d are
c. For the reaction 2. CACO,(s) + 2H+ (aq)→ Ca (ag) + H,0(1) + CO, Species S J(mol K) CACO3 (s) 92.9 H* (ag) 0. Ca (aq) -53.1 H20(1) 69.95 CO, (9) 213.7 AS = JK
Given the data in the table below, AHrxn for the reaction Ca(OH); + 2H;AsO4 → Ca(H;AsO4)2 + 2H;O is _ kJ. [A] -4219 (B] -130.4 [C] -4519 [D] -76.4 AH, (kJ/mol) -986.6|| Substance Ca(ОН), HASO, Ca(H;AsOa)2 -900.4 -2346.0 H;0 -285.9 O A В C D
Consider a general reaction enzyme A(aq) = B(aq) The AGo of the reaction is-8.290 kJ · mol. Calculate the equilibrium constant for the reaction at 25 °C. %3D What is AG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.7 M and the concentration of B is 0.50 M? kJ mol-! AG = %3D |help terms of use contact us careers privacy policy about s MacBook Pro
Help!!
Can you help me with this?
36