Cu metal crystallizes with fcc structure. The density of crystal is 2698 kg/m3 and mass is 26.98 Kg. X-ray of wavelength 1.537 nm are diffracted by the crystal and the Bragg’s angle is 19.20. Identify the crystal plane (m,n,o) which caused this diffraction and give the value (m + n + o):Correct answer is '3'. Can you explain this answer?

Question

Cu metal crystallizes with fcc structure. The density of crystal is 2698 kg/m3 and mass is 26.98 Kg. X-ray of wavelength 1.537 nm are diffracted by the crystal and the Bragg&rsquo;s angle is 19.20. Identify the crystal plane (m,n,o) which caused this diffraction and give the value (m + n + o):Correct answer is '3'. Can you explain this answer?

Answer

The Bragg's Law equation relates the wavelength of X-rays diffracted by a crystal to the interplanar spacing and the angle of diffraction.

The equation is given as:

nλ = 2d sinθ

Where:
n = order of diffraction (usually 1 for X-rays)
λ = wavelength of X-rays (1.537 nm = 1.537 x 10^-9 m)
d = interplanar spacing
θ = angle of diffraction

To find the interplanar spacing, we need to use the formula for the fcc crystal structure:

d = a/√(h^2 + k^2 + l^2)

Where:
a = lattice parameter (side length of the unit cell)
h, k, l = Miller indices (indices of the crystallographic plane)

The fcc crystal structure has lattice points at the corners and face centers of the unit cell. Therefore, the Miller indices for the (hkl) plane in the fcc structure are: (hkl) = (hkl).

Given that the density of the crystal is 2698 kg/m3 and the mass is 26.98 kg, we can find the volume of the crystal using the density formula:

Density = Mass / Volume

Volume = Mass / Density
= 26.98 kg / 2698 kg/m3
= 0.01 m3

Since the fcc unit cell contains 4 atoms, the volume of the unit cell is 4 times the volume of the crystal:

Volume of unit cell = 4 x Volume
= 4 x 0.01 m3
= 0.04 m3

The volume of the unit cell can also be expressed in terms of the lattice parameter (a):

Volume of unit cell = a^3

Therefore, a^3 = 0.04 m3

Taking the cube root of both sides, we find:

a = ∛(0.04 m3)
= 0.33 m

Now, we can calculate the interplanar spacing using the formula for the fcc structure:

d = a / √(h^2 + k^2 + l^2)
= 0.33 m / √(h^2 + k^2 + l^2)

Since we don't have any specific planes given in the question, we can assume the (111) plane:

d = 0.33 m / √(1^2 + 1^2 + 1^2)
= 0.33 m / √3
= 0.19 m

Now, we can use Bragg's Law to find the angle of diffraction:

nλ = 2d sinθ

Rearranging the equation to solve for sinθ, we get:

sinθ = (nλ) / (2d)
= (1 x 1.537 x 10^-9 m) / (2 x 0.19 m)
= 4.04 x 10^-9

Taking the inverse sine of sinθ, we find:

θ = sin^(-1)(4.04 x 10^-9)
= 2.32 x 10^-7 radians

Finally, we convert the angle to degrees:

θ (in degrees) = 2.32 x 10

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