Calculate the separation of (a) (123) plane and (b) (246) plane of an orthorombic cell with a =0.82nm,b=0.94nm and c =0.75nm

Answer 1

To calculate the separation of a specific plane in an orthorhombic crystal structure, we can use the formula:

Separation distance = 1 / ( sqrt( h^2 / a^2 + k^2 / b^2 + l^2 / c^2 ) )

where a, b, and c are the cell parameters, and h, k, and l are the Miller indices of the plane.

For plane (123), h=1, k=2, and l=3. Plugging in the values:

Separation distance (123) = 1 / (sqrt(1^2 / 0.82^2 + 2^2 / 0.94^2 + 3^2 / 0.75^2))

Separation distance (123) = 1 / (sqrt(1.22 + 4.54 + 14.4))

Separation distance (123) = 1 / (sqrt(20.16))

Separation distance (123) = 1 / 4.49

Separation distance (123) = 0.223 nm

For plane (246), h=2, k=4, and l=6. Plugging in the values:

Separation distance (246) = 1 / (sqrt(2^2 / 0.82^2 + 4^2 / 0.94^2 + 6^2 / 0.75^2))

Separation distance (246) = 1 / (sqrt(4.88 + 16.28 + 28.8))

Separation distance (246) = 1 / (sqrt(49.96))

Separation distance (246) = 1 / 7.07

Separation distance (246) = 0.141 nm

Therefore, the separation of the (123) plane is 0.223 nm and the separation of the (246) plane is 0.141 nm in the given orthorhombic cell.