6. A careless person has disposed of arsenic in a well used for drinking water. The arsenic sits at the bottomof the well and diffuses into the water. This diffusion obeys the following equation:Yt = DYxx (1)where Y is the concentration of arsenic. The surface of the well is at x = 0 and the bottom at x = L =20m. At x = 0 the concentration is fixed at Y = 0 and at x = L the concentration is fixed at Y = 1.(a) [2]What is the steady state concentration of arsenic in the well?Solution: The steady-state condition is:Yt = 0Therefore the governing equation becomes:0 = DYxxIntegrating twice:0 = DY + c1x + c2Where c1 and c2 are constants of integration. Substituting the boundary conditions Y (0, t) = 0and Y (L, t) = 1:Y = xL(b) The arsenic is removed from the bottom after the concentration has reached a steady state. Unpol-luted ground water now keeps the concentration at x = L fixed at Y = 0 while the concentration atx = 0 remains fixed at Y = 0. You need to find how the concentration of arsenic changes in time.i. [2]Sketch the initial and boundary conditions.Solution: The initial and boundary conditions look like this

For a 𝑝-type substrate doped to NA = 4.5x1015 cm-3 and diffusion of arsenic from a constantsurface density of 1.5x1021 cm-3, calculate (use k = 8.62x10-5 eV/K):(a) the junction depth 𝑥𝑗 (in m) for diffusion at T = 1100 °C for t = 90 min. [10 Pts](b) the time (in min) required for a junction depth of 𝑥𝑗= 1 m at 1100 °C. [10 Pts]Use the diffusion equation below:𝑁(𝑥, 𝑡) = 𝑁0𝑒𝑟𝑓𝑐 ( 𝑥2√𝐷𝑡)where𝑁0 is the surface density (#/cm3).𝐷 is the diffusion coefficient and given by 𝐷 = 24exp(-Ea/kT) cm2/s for arsenicin silicon. Ea = 4.08 eV.𝑡 is the length of time the diffusion takes place (s)The complementary error function erfc(y) can be found in many math apps includingExcel

A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.7kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. Answer the following questions.1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t).Answer (in kilograms per minute): dydt =

A tank is filled with 1000 liters of pure water. Brine containing 0.04 kg of salt per liter enters the tank at 6 liters per minute. Another brine solution containing 0.06 kg of salt per liter enters the tank at 6 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 12 liters per minute.A. Determine the differential equation which describes this system. Let 𝑆(𝑡) denote the number of kg of salt in the tank after 𝑡 minutes.

(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.7kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. Answer the following questions.1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t).Answer (in kilograms per minute): dydt =

Arsenic is ion implanted into a uniformly doped, bare 𝑝-type Si substrate doped to NA =1016 cm-3. The implant energy is 𝐸 = 100 keV and the implant dose is 𝜙 = 8x1014 cm-2. For aprojected range of 𝑅𝑝 = 0.06 m and a projected straggle of Δ𝑅𝑝= 0.02 m, calculate:(a) the peak As density (in cm-3) [10 Pts](b) the surface As density (in cm-3) [10 Pts](c) the junction depth 𝑥𝑗 (in m). [10 Pts]𝑁(𝑥) = 𝜙√2𝜋∆𝑅𝑝𝑒𝑥𝑝 (− (𝑥 − 𝑅𝑝)22∆𝑅𝑝2