Question

lead has fcc structure with lattice constant of 4.95 A. calculate interplanar spacing of (110) plane
À. 2.86 Å
b.3.5 Å
c. 4.24 Å
d. 7.49 Å​

Answer 1

Answer:

7.49 ok ji aap ki baat nhi

Explanation:

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Answer 2

Answer:

Lead has fcc structure with a lattice constant of 4.95 A. The interplanar spacing of (110) plane will be $3.500$ Å

Explanation:

The d- spacing or the distance between the consecutive lattice planes of a Face centered cubic lattice can be expressed as follows,

$d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$          ...(1)

Where,

h,k,l    -  Miller indices of the adjacent planes  to be considered

a         - lattice constant

Given,

( hkl )  = ($110$)

h = 1

k = 1

l  = 0

Lattice constant a = $4.95$ Å

Equation (1) becomes,

$d_{110}=\frac{4.95 \AA}{\sqrt{1^{2}+1^{2}+0^{2}}}$

       $=\frac{4.95}{\sqrt{2} } \AA= 3.500\ \AA$

The correct answer is option b.