Very simple vector question (solid state)

AI Thread Summary
The discussion revolves around calculating the carbon-carbon bond length in graphene, specifically deriving the vector magnitude |r2|, which is found to be 1.42 Å. The calculations involve using the lattice vectors a and b, with magnitudes of 2.46 Å and an angle of 60° between them. Participants clarify that the magnitude of a vector can be determined using the scalar product and the cosine rule, emphasizing the correct application of these concepts. Confusion arises regarding the formulas used, with some suggesting alternative approaches to derive |r2|. Ultimately, the key takeaway is understanding how to apply geometric principles to solve for the bond length in the context of graphene's crystal structure.
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Homework Statement
Graphene is a 2D crystal made entirely of carbon atoms. The crystal lattice can defined by the lattice vectors a and b, where the magnitudes |a|=|b|= 2.46 Å and the angle between the lattice vectors is 60°. The crystal has a basis of two carbon atoms, where the basis vectors are r1 = 0 and r2 = a/3 + b/3.

Sketch the unit cell of the crystal structure of graphene and calculate the carbon- carbon bond length (in Å).
Relevant Equations
Bond length = |r2|
Screenshot 2019-07-20 at 22.40.36.png


Somewhat embarrassingly as a third year undergrad, this question has been completely stumping me for far too long now (2 hours). The solution is 1.42 Å and the working is given as

|r2| = 2cos(30)*1/3(2.46)

or alternatively

|r2| = (1/2|a|)/cos(30)

But I cannot grasp where this comes from. Please someone put me out of my misery :sorry:
 
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Phys128 said:
Problem Statement: Graphene is a 2D crystal made entirely of carbon atoms. The crystal lattice can defined by the lattice vectors a and b, where the magnitudes |a|=|b|= 2.46 Å and the angle between the lattice vectors is 60°. The crystal has a basis of two carbon atoms, where the basis vectors are r1 = 0 and r2 = a/3 + b/3.

Sketch the unit cell of the crystal structure of graphene and calculate the carbon- carbon bond length (in Å).
Relevant Equations: Bond length = |r2|

View attachment 246904

Somewhat embarrassingly as a third year undergrad, this question has been completely stumping me for far too long now (2 hours). The solution is 1.42 Å and the working is given as

|r2| = 2cos(30)*1/3(2.46)

or alternatively

|r2| = (1/2|a|)/cos(30)

But I cannot grasp where this comes from. Please someone put me out of my misery :sorry:
You need the magnitude of the vector ##\vec r_2##. How is the magnitude of a vector defined, using scalar product?
 
ehild said:
You need the magnitude of the vector ##\vec r_2##. How is the magnitude of a vector defined, using scalar product?

|r2| = |a||b|cosx but then should that not be [(2.46/3)^2]cos(30) rather than 2(2.46/3)cos30?
 
Phys128 said:
|r2| = |a||b|cosx but then should that not be [(2.46/3)^2]cos(30) rather than 2(2.46/3)cos30?

No, your formula would givse the scalar product of two vectors a and b, inclined at angle x.
The magnitude of a vector is defined as the square root of its scalar product by itself.
Or you can use the cosine rule to calculate the length of r2, but remember, the angle between a and b is 60°.
 
The comments of @ehild are great.

The following is not much different than her comments. But, maybe it will be helpful.
Phys128 said:
The solution is 1.42 Å and the working is given as

|r2| = 2cos(30)*1/3(2.46)
In order to see this, consider an isosceles triangle as shown:
P1.png

Show that ##z = 2x\cos\theta##. Apply this to your situation.
or alternatively

|r2| = (1/2|a|)/cos(30)

To see this way of expressing it, consider the right triangle shown in the figure below:
P2.png

Show that the horizontal leg of the triangle has length ##a/2## as indicated. Once you have that, the result follows quickly.
 
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