The lattice constant of a single crystal is 4.73 Å. Calculate the surface density (# / cm^2) of

First, we need to convert the lattice constant from angstroms to centimeters: $4.73 AA = 4.73

The lattice constant of a single crystal is 4.73 Å....

The lattice constant of a single crystal is $4.73 \AA$. Calculate the surface density $\left(\# / \mathrm{cm}^{2}\right)$ of atoms on the (i) (100), (ii) (110), and (iii) (111) plane for a ( $a$ ) simple cubic, (b) body-centered cubic, and $(c)$ face-centered cubic lattice.

Transcript

00:01 Hello students, you are given that a equal to 4 .73 and the plane it is given that 1 comma 0 0 .0 for plane 1 comma 0 area equal to a square which is equals to the 4 .73 square equals to the value of 22 .372 now we can calculate the surface density for simple cube and body centered cube and face center cube for simple cube n is equal to one therefore it is given that surface density row is equal to n by a for any plane to calculate the surface density this is this is the formula therefore row equal to 1 by area a equal to a square we have calculated 1 by 22 .372 which is equals to the value of row equal to this 0 .0 .04 angstroms therefore row equal to 0 .044 angstroms now for body centered cube for body centered cube number of n is equal to 2 therefore row equal to 2 by area a equal to 22 .372 which is equals to the value of 0 .089 therefore row equal to 0 .0 894 angstroms.

01:40 For part c, for fcc, n is equal to 4.

01:46 Therefore, the obtain value of row equal to 4 by 22 .372, which is equals to the value of 0 .178.

01:59 Therefore, row for fcc is given as 0 .177.

02:04 178 for 1 comma 0 comma 0 now for second it is given that 1 comma 1 comma 0 for 1 comma 0 area equals to root 2 into a square given that a equal to 4 .73 square which is equal to the value of 31 .64 area for simple cube again for simple cube cube cube n equal to 1 therefore row equal to n by area a which is equal to 1 by area is calculated 31 .64 which is equal to the value of row equal to 0 .013 0 .036 angstroms for b c c d is given that for b c c for 1 .0 .0 .0 n equal to 2 for row equal to n by a equals to 2 by 31 .64 therefore the obtained value of row equals to 0 .0 632 angstroms now for it is given that for the value of fcc 1 comma 0 comma 0 n is equal to 4 row equal to 4 by 31 .64 therefore row is equal to 0 .126 angstroms.

03:50 Now we can calculate for the plane 1 comma 1 comma 1 .1 the third plane it is given that 1 comma 1 here 1 comma 1 area is given as root 3 by 2 into a square which is equal to root 3 by 2 into 4 .73 square which is equal to root 3 by 2 into 4 .73 square equals to the value of 19 .37 area.

04:22 Now we can calculate the density, surface density, which is equal to for simple cube, per simple cube, n is equal to 1 and area is calculated 19 .37.

04:40 Therefore, row is equal to n by a equal to 1 by 19 .37...

Question

The lattice constant of a single crystal is $4.73 \AA$. Calculate the surface density $\left(\# / \mathrm{cm}^{2}\right)$ of atoms on the (i) (100), (ii) (110), and (iii) (111) plane for a ( $a$ ) simple cubic, (b) body-centered cubic, and $(c)$ face-centered cubic lattice.

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