UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper I

Q5(d). Spacing of the (111) planes of a simple cubic lattice

Question

What is the shortest distance between two successive 111 planes of simple cubic lattice of edge length 200 pm? Explain.

d₁₁₁=a/√3=200/√3=115.5 pm

The (111) plane cuts the three axes at a, a, a (reciprocals 1, 1, 1). In a cubic lattice the spacing of the hkl planes is dₕₖₗ=ah²+k²+l².
d₁₁₁=(200 pm)/(√(1+1+1))=(200)/(1.732)=115.5 pm
the (111) planes are x+y+z=0, a, 2a, 3a. The body diagonal from (0,0,0) to (a,a,a) is normal to them and has length √3a=346 pm. It meets the planes at 0, 1/3, 2/3 and 1 of its length, so successive planes are √3a/3=a/√3 apart.
Every lattice point lies on one of these planes, so none lies in between: 115.5 pm is the shortest distance. (d₁₀₀=200, d₁₁₀=141.4 pm: higher indices, closer planes.)

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

The (111) plane cuts the three axes at a, a, a (reciprocals 1, 1, 1). In a cubic lattice the spacing of the hkl planes is dₕₖₗ=ah²+k²+l².
d₁₁₁=(200 pm)/(√(1+1+1))=(200)/(1.732)=115.5 pm
Explanation: the (111) planes are x+y+z=0, a, 2a, 3a. The body diagonal from (0,0,0) to (a,a,a) is normal to them and has length √3a=346 pm. It meets the planes at 0, 1/3, 2/3 and 1 of its length, so successive planes are √3a/3=a/√3 apart.
Every lattice point lies on one of these planes, so none lies in between: 115.5 pm is the shortest distance. (d₁₀₀=200, d₁₁₀=141.4 pm: higher indices, closer planes.)

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