Cubic systems — SCC, BCC, FCC; packing fraction calculations

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WHAT is a unit cell?

WHY count atoms as fractions? Because atoms sit on the shared boundaries between neighbouring boxes. An atom at a corner is shared by 8 boxes, so this box only "owns" $1/8$ of it.

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The three simple cubic types

1. Simple Cubic (SCC)

Atoms only at the 8 corners.

$\phi_{SCC}=\frac{Z\cdot\frac43\pi r^3}{a^3}=\frac{1\cdot\frac43\pi r^3}{(2r)^3}=\frac{\frac43\pi r^3}{8r^3}=\frac{\pi}{6}\approx ==0.524== \;(52.4\%)$

2. Body-Centred Cubic (BCC)

8 corners + 1 atom at the body centre.

Derive the body diagonal. For a cube of edge $a$: face diagonal $=\sqrt{a^2+a^2}=a\sqrt2$; body diagonal $=\sqrt{a^2+(a\sqrt2)^2}=\sqrt{3a^2}=a\sqrt3$. So $a\sqrt3 = 4r \Rightarrow a=\dfrac{4r}{\sqrt3}$.

$\phi_{BCC}=\frac{2\cdot\frac43\pi r^3}{\left(\frac{4r}{\sqrt3}\right)^3}=\frac{\frac83\pi r^3}{\frac{64r^3}{3\sqrt3}}=\frac{8\pi\cdot 3\sqrt3}{3\cdot 64}=\frac{\pi\sqrt3}{8}\approx ==0.680== \;(68\%)$

3. Face-Centred Cubic (FCC / CCP)

8 corners + 6 face centres.

$a\sqrt2 = 4r \Rightarrow a=\dfrac{4r}{\sqrt2}=2\sqrt2\,r$.

$\phi_{FCC}=\frac{4\cdot\frac43\pi r^3}{(2\sqrt2 r)^3}=\frac{\frac{16}{3}\pi r^3}{16\sqrt2\,r^3}=\frac{\pi}{3\sqrt2}=\frac{\pi\sqrt2}{6}\approx ==0.740== \;(74\%)$

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Density link (why we care in exams)

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Worked examples

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Common mistakes

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Flashcards

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Recall Feynman: explain to a 12-year-old

Imagine packing oranges into a box. If you just stack them straight in rows (simple cubic), lots of gaps stay empty — only about half the box is fruit. If you slip one orange into the middle gap (body-centred), it fits tighter, 68%. If you nestle oranges into the dimples of the layer below (face-centred), it's the tightest possible, 74%. The "packing fraction" is just: how much of the box is orange, not air. The trick is figuring out which oranges are actually touching so you know how big they can be for a given box.

Connections

• Coordination Number in Crystals (SCC=6, BCC=8, FCC=12)
• Voids — Tetrahedral and Octahedral
• Density of a Unit Cell
• Pythagoras Theorem (used for face & body diagonals)
• Close Packing — HCP vs CCP
• Avogadro Number and Molar Mass

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, crystal ek dabbe (unit cell) mein atoms ko sphere maan ke stack karte hain. Sabse important sawaal hai: dabbe ka kitna hissa atom se bhara hai aur kitna khali? Isi ko packing fraction kehte hain. Iske liye teen cheez chahiye — Z (kitne pure atom ek box ke andar aate hain), $a$ aur $r$ ka relation, aur phir simple division: (atoms ka volume)/(cube ka volume).

Corner atom 8 dabbon mein share hota hai, isliye ek box ko sirf $1/8$ milta hai; face wala $1/2$, body-centre wala poora $1$. Isse Z nikalta hai — SCC mein 1, BCC mein 2, FCC mein 4. Ab sabse bada trick: atoms kis direction mein touch karte hain? SCC mein edge ke along ($a=2r$), BCC mein body diagonal ke along ($\sqrt3 a=4r$), FCC mein face diagonal ke along ($\sqrt2 a=4r$). Bas yahi yaad rakho, baaki sab derive ho jayega.

In values ko lagate hi packing fractions milte hain: SCC $=52.4\%$, BCC $=68\%$, FCC $=74\%$. Matlab jitne zyada atoms, utni tight packing. FCC sabse dense hai — isiliye ise "closest packing" bolte hain, sirf 26% khali. Aur density ke sawaal ke liye formula $\rho = ZM/(a^3 N_A)$ hai — bas dhyaan rakho $a$ ko cm mein convert karna, warna answer $10^{30}$ times galat aayega. Yeh JEE/NEET ka favourite trap hai!