Question

Problem 1. An image of graphene h obtained using High Resolution Transmission Electron Microscopy (HRTEM) is shown on Fig. 1. The length of the carbon bonds is a = 0.142nm. 1. Calculate the density of carbon atoms (at/cm²) 2. Given the vectors of the direct lattice $\vec{a_1}$ and $\vec{a_2}$ calculate the reciprocal lattice vectors $\vec{b_1}$ and $\vec{b_2}$ defined as follows: $\vec{b_1} = \frac{2\pi}{A} \vec{a_2} \times \vec{n}$ and $\vec{b_2} = \frac{2\pi}{A} \vec{n} \times \vec{a_2}$ where A = $|\vec{a_1} \times \vec{a_2}|$ and $\vec{n}$ (0,0,1) 3. Draw the first Brillouin zone and Calculate its area. 4. Calculate the distance ?M between the center of BZ and the edge of the BZ. and the distance ?K to any ocrner of the Brillouin Zone.

          Problem 1.
An image of graphene h obtained using High Resolution
Transmission Electron Microscopy (HRTEM) is shown on Fig. 1.
The length of the carbon bonds is a = 0.142nm.
1. Calculate the density of carbon atoms (at/cm²)
2. Given the vectors of the direct lattice $\vec{a_1}$ and $\vec{a_2}$ calculate
the reciprocal lattice vectors $\vec{b_1}$ and $\vec{b_2}$ defined as
follows:
$\vec{b_1} = \frac{2\pi}{A} \vec{a_2} \times \vec{n}$ and $\vec{b_2} = \frac{2\pi}{A} \vec{n} \times \vec{a_2}$ where A = $|\vec{a_1} \times \vec{a_2}|$ and $\vec{n}$ (0,0,1)
3. Draw the first Brillouin zone and Calculate its area.
4. Calculate the distance ?M between the center of BZ and the edge of the BZ.
and the distance ?K to any ocrner of the Brillouin Zone.

Show more…

Problem 1.
An image of graphene h obtained using High Resolution
Transmission Electron Microscopy (HRTEM) is shown on Fig. 1.
The length of the carbon bonds is a = 0.142nm.
1. Calculate the density of carbon atoms (at/cm²)
2. Given the vectors of the direct lattice a⃗1⃗ and a⃗2⃗ calculate
the reciprocal lattice vectors b⃗1⃗ and b⃗2⃗ defined as
follows:
b⃗1⃗ = (2π)/(A)a⃗2⃗×n⃗ and b⃗2⃗ = (2π)/(A)n⃗×a⃗2⃗ where A = |a⃗1⃗×a⃗2⃗| and n⃗ (0,0,1)
3. Draw the first Brillouin zone and Calculate its area.
4. Calculate the distance ?M between the center of BZ and the edge of the BZ.
and the distance ?K to any ocrner of the Brillouin Zone.

Added by Lisa W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Susan Hallstrom

12/26/2023

Step 1

The area of the unit cell can be calculated using the formula A = |vec(a)_(1) × vec(a)_(2)|. Once we have the area of the unit cell, we can calculate the density of carbon atoms by dividing the number of carbon atoms in the unit cell by the area of the unit cell. Show more…

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Problem 1: An image of graphene obtained using High Resolution Transmission Electron Microscopy (HRTEM) is shown in Fig. 1. The length of the carbon bonds is a = 0.142 nm. 1. Calculate the density of carbon atoms (atoms/cm^2). 2. Given the vectors of the direct lattice vec(a)_(1) and vec(a)_(2), calculate the reciprocal lattice vectors vec(b)_(1) and vec(b)_(2) defined as follows: vec(b)_(1) = (2π)/(A) vec(a)_(2) × vec(n) and vec(b)_(2) = (2π)/(A) vec(n) × vec(a)_(2) where A = |vec(a)_(1) × vec(a)_(2)| and vec(n) = (0,0,1). 3. Draw the first Brillouin zone and calculate its area. 4. Calculate the distance ΓM between the center of BZ and the edge of the BZ, and the distance ΓK to any corner of the Brillouin Zone.

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Transcript

00:01 In this problem we're asked to consider graphene.

00:05 And i've attached an image here.

00:08 It's a sheet of carbon atoms, blah, blah, blah.

00:20 It can be considered as an extreme case of polyaromatic hydrocarbon of infinite length, blah, blah.

00:30 And here is, this is the image i want to consider right here.

00:38 So we are asked the area of one hexagonal six carbon unit is approximately 5 ,000, or 52 ,400 picometers squared.

01:13 Calculate the number of pi bonds in a 25 nanometer by 25 nanometer sheet of graphene.

01:44 And we are told that we can ignore edge electrons.

01:55 Okay, so you can see my hexagonal units.

01:59 I'll make one a little bit bigger here.

02:02 Is like, okay, number of pi bonds.

02:33 So i'm gonna get over to my book here for a sec.

02:38 One, two, three, four, five, six carbon unit.

03:05 So my sheet area, i'm gonna switch colors here, is 25 times 25 nanometers.

03:20 That'll equal 625, let's see, nanometers squared.

03:42 And one carbon, one hexagonal carbon unit is 52 ,400.

04:01 Picometers and picometers squared.

04:08 And there'll be 100, 1 ,000 picometers squared per one nanometer will be, let me see, 5 ,2 ,4 ,0 ,0 divided by 1 ,000, 1 ,000 is zero.

04:34 Let me do this again.

04:36 0 .0524 nanometers squared.

04:49 Okay, i forgot that i can't scroll down on this one.

04:56 If i'm gonna have enough room here.

04:58 So now i have 625 nanometers squared divided by 0 .0524 nanometers squared...

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