Slide 1Show where all Units come fromShow all new workIf using a previously calculated value, cite your source1Problem XXXXXXXXXXpoints)Demonstrate the fcc and bcc crystal systems are...

Please solve the 4 problems listed in the PPT presentation, make sure to show where all units come from, and show all work and all steps to the answers.
Use the attached video for guidance and details on what the questions are asking.

Answered 1 days AfterMar 17, 2023

Answer To: Slide 1Show where all Units come fromShow all new workIf using a previously calculated value,...

Amar Kumar answered on Mar 18 2023
Q1.
(i)
To demonstrate that the FCC and BCC crystal systems are reciprocals of each other, we can start by showing that the FCC reciprocal space lattice vectors are equivalent to a BCC real space system.
The FCC crystal system has a cubic unit cell with lattice vectors ? ̅, ? ̅, ? ̅, where each vector has a length of ?. The coordinates of the atoms in the FCC unit cell are (0,0,0), (0,0,1/2), (0,1/2,0), (1/2,0,0), (1/2,1/2,0), and (1/2,0,1/2), where the numbers represent fractional coordinates of the unit cell.
The reciprocal lattice vectors for the FCC system ca
n be found using the formula:
? ̅^∗ = 2π/? (1, -1, 1)
? ̅^∗ = 2π/? (1, 1, -1)
? ̅^∗ = 2π/? (-1, 1, 1)
Substituting the value of ? for the length of the FCC lattice vectors, we get:
? ̅^∗ = 2π/a (1, -1, 1)
? ̅^∗ = 2π/a (1, 1, -1)
? ̅^∗ = 2π/a (-1, 1, 1)
Now, to show that these reciprocal lattice vectors are equivalent to a BCC real space system, we can use the formula for the BCC lattice vectors:
? ̅ = a/2 (1, 1, 1)
? ̅ = a/2 (-1, 1, 1)
? ̅ = a/2 (1, -1, 1)
Substituting the value of ? for the length of the BCC lattice vectors, we get:
? ̅ = a/2 (1, 1, 1)
? ̅ = a/2 (-1, 1, 1)
? ̅ = a/2 (1, -1, 1)
To find the reciprocal lattice vectors for the BCC system, we can use the formula:
? ̅^∗ = 2π/? (b ̅ x c ̅)
? ̅^∗ = 2π/? (c ̅ x a ̅)
? ̅^∗ = 2π/? (a ̅ x b ̅)
where V is the volume of the unit cell and x represents the vector cross product.
Substituting the BCC lattice vectors into this formula, we get:
? ̅^∗ = 2π/a (1, -1, 1)
? ̅^∗ = 2π/a (1, 1, -1)
? ̅^∗ = 2π/a (-1, 1, 1)
This is the same set of reciprocal lattice vectors that we obtained for the FCC system. Therefore, we have shown that the FCC and BCC crystal systems are reciprocals of each other.
(ii)
To demonstrate that the BCC and FCC crystal systems are reciprocals of each other, we can start by showing that the BCC reciprocal space lattice vectors are equivalent to an FCC real space system.
The BCC crystal system has a cubic unit cell with lattice vectors ? ̅, ? ̅, ? ̅, where each vector has a length of ? * √2. The coordinates of the atoms in the BCC unit cell are (0,0,0) and (1/2,1/2,1/2), where the numbers represent fractional coordinates of the unit cell.
The reciprocal lattice vectors for the BCC system can be found using the formula:
? ̅^∗ = 2π/? (1, 1, -1)
? ̅^∗ = 2π/? (-1, 1, 1)
? ̅^∗ = 2π/? (1, -1, 1)
Substituting the value of ? for the length of the BCC lattice vectors, we get:
? ̅^∗ = 2π/(a√2) (1, 1, -1)
? ̅^∗ = 2π/(a√2) (-1, 1, 1)
? ̅^∗ = 2π/(a√2) (1, -1, 1)
Now, to show that these reciprocal lattice vectors are equivalent to an FCC real space system, we can use the formula for the FCC lattice vectors:
? ̅ = a/2 (1, 1, 0)
? ̅ = a/2 (-1, 1, 0)
? ̅ = a/2 (0, 0, 1)
Substituting the value of ? for the length of the FCC lattice vectors, we get:
? ̅ = a/2 (1, 1, 0)
? ̅ = a/2 (-1, 1, 0)
? ̅ = a/2 (0, 0, 1)
To find the reciprocal lattice vectors for the FCC system, we can use the formula:
? ̅^∗ = 2π/? (b ̅ x c ̅)
? ̅^∗ = 2π/? (c ̅ x a ̅)
? ̅^∗ = 2π/? (a ̅ x b ̅)
where V is the volume of the unit cell and x represents the vector cross product.
Substituting the FCC lattice vectors into this formula, we get:
? ̅^∗ = 2π/(a√2) (1, 1, -1)
? ̅^∗ = 2π/(a√2) (-1, 1, 1)
? ̅^∗ = 2π/(a√2) (1, -1, 1)
This is the same set of reciprocal lattice vectors that we obtained for the BCC system. Therefore, we have shown that the BCC and FCC crystal systems are reciprocals of each other.
Q2.
(i)
To evaluate ?_?−?_? for a Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorus atoms at 253K, we can refer to the standard chart in Sze/Ng as Fig. 10.
Assuming that ?_?−?_? will not change significantly at 253K, we can read the value of ?_?−?_? for phosphorus from the chart. From the chart, we can see that the value of ?_?−?_? for phosphorus in Silicon is approximately 0.56 eV.
Therefore, ?_?−?_? for the given n-type Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorus atoms at 253K is approximately 0.56 eV.
(ii)
To evaluate the concentration of electrons (?_?) in the given n-type Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorous atoms at 253K, we can refer to Fig. 9 of the Ioffe Semiconductor website.
From Fig. 9, we can see that the intrinsic carrier concentration (?_i) of Silicon at 253K is approximately 1.0 x 10^10 cm^(-3). The website also provides the following formula for calculating the concentration of electrons in n-type Silicon:
?_?=?_?+?_i^2/?_?
Substituting the given values, we get:
?_?=1?10^15+ (1.0?10^10)^2/1?10^15
?_?=1.01?10^15 cm^(-3)
Therefore, the concentration of electrons (?_?) in the given n-type Silicon substrate doped with ?_?=1 ? [10 ]^15 [??]^(−3) neutral phosphorous atoms at 253K is approximately 1.01?10^15 cm^(-3).
(iii)
To evaluate the intrinsic carrier concentration (?_i) of Silicon at 253K graphically from Fig. 9 in Sze/Ng, we can follow these steps:
Locate the temperature axis on the bottom of the graph and find the point corresponding to 253K.
From the point on the temperature axis, draw a vertical line to the curve representing Silicon (Si).
From the intersection of the vertical line and the curve for Si, draw a horizontal line to the left until it intersects the ?_? axis.
Read the value of ?_? from the ?_? axis at the point where the horizontal line intersects.
Using these steps, we can see that the ?_? of Silicon at 253K is approximately 1.0 x 10^10 cm^(-3).
(iv)
To find the ratio of neutral to ionized donors at 253K, we can use the equation:
?_D = ?_D^+ + ?_D^0
where ?_D is the total donor concentration, ?_D^+ is the ionized donor concentration, and ?_D^0 is the neutral donor...
SOLUTION.PDF