Q1(d)(ii). Spacing of the (110) planes, axes mutually perpendicular

Answer in brief

(1)/(d₁₁₀²)=(1)/(a²)+(1)/(b²), so d₁₁₀=(ab)/(√(a²+b²)); for a=b (cubic, tetragonal) d₁₁₀=a/√2.

Key points

• For a lattice with mutually perpendicular axes (orthorhombic, and as special cases tetragonal and cubic): (1)/(dₕₖₗ²)=(h²)/(a²)+(k²)/(b²)+(l²)/(c²)
• For (110), h=k=1, l=0: (1)/(d₁₁₀²)=(1)/(a²)+(1)/(b²)=(a²+b²)/(a²b²), d₁₁₀=(ab)/(√(a²+b²)) (c does not enter)
• Geometric check. The (110) plane cuts x at a, y at b and runs parallel to c. Seen along c it is the line joining (a,0) and (0,b). Its distance from the origin is twice the triangle area divided by the hypotenuse: ab/√(a²+b²).
• Tetragonal and cubic (a=b): d₁₁₀=a/√2=0.7071 a; for a=400 pm, d₁₁₀=282.8 pm. For an orthorhombic cell with a=500, b=600 pm, d₁₁₀=384.1 pm.