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NCERT Exemplar · Q11

Q.Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N2\text{N}_2 (∼5×10−9\sim 5 \times 10^{-9}) (at STP) and Cu (∼10−5\sim 10^{-5}).

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The huge difference arises because diamagnetic susceptibility depends on the number density of atoms and the size of the electron orbits. In a gas like N₂ at STP, atoms are far apart (low density), while in a solid metal like Cu, atoms are tightly packed (high density). Additionally, copper has more electrons per atom and larger effective orbital radii, giving a much larger induced magnetic moment per atom. The combined effect yields a factor of about 10410^4 — exactly the observed gap.

Why this approach works

Diamagnetism is a universal property: when an external magnetic field is applied, it slightly alters the orbital motion of electrons, inducing a tiny magnetic moment that opposes the field. The size of this induced moment per atom is proportional to the square of the orbital radius and the number of electrons. But the bulk susceptibility χ\chi also depends on how many atoms are packed into a given volume — the number density.

So the order-of-magnitude difference between N₂ gas and solid Cu comes from two separate factors:

  1. Number density — how many atoms per cubic metre.
  2. Atomic diamagnetic response — how large the induced moment is per atom.

Let’s quantify each.

Step-by-step calculation

1. Number density at STP vs. in a solid

For an ideal gas at STP (0 °C, 1 atm), one mole occupies 22.4 L = 2.24×10−2 m32.24 \times 10^{-2} \,\text{m}^3.

Number of molecules per mole is Avogadro’s number NA=6.02×1023N_A = 6.02 \times 10^{23}.

So number density of N₂ molecules:

nN2=6.02×10232.24×10−2≈2.69×1025 m−3n_{\text{N}_2} = \frac{6.02 \times 10^{23}}{2.24 \times 10^{-2}} \approx 2.69 \times 10^{25} \,\text{m}^{-3}

For copper: density ρ=8.96 g/cm3=8960 kg/m3\rho = 8.96 \,\text{g/cm}^3 = 8960 \,\text{kg/m}^3, atomic mass M=63.5 g/mol=0.0635 kg/molM = 63.5 \,\text{g/mol} = 0.0635 \,\text{kg/mol}.

Number density of Cu atoms:

nCu=ρNAM=8960×6.02×10230.0635≈8.5×1028 m−3n_{\text{Cu}} = \frac{\rho N_A}{M} = \frac{8960 \times 6.02 \times 10^{23}}{0.0635} \approx 8.5 \times 10^{28} \,\text{m}^{-3}

Ratio of number densities:

nCunN2≈8.5×10282.69×1025≈3.2×103\frac{n_{\text{Cu}}}{n_{\text{N}_2}} \approx \frac{8.5 \times 10^{28}}{2.69 \times 10^{25}} \approx 3.2 \times 10^{3}

So just from packing, Cu has about 3000 times more atoms per unit volume than N₂ gas.

Tip

This factor alone already accounts for most of the difference — but not all. The remaining factor comes from the atomic diamagnetic response.

2. Atomic diamagnetic susceptibility per atom

The classical Langevin formula for diamagnetic susceptibility per atom (or molecule) is:

χatom=−μ0e26me∑⟨r2⟩\chi_{\text{atom}} = -\frac{\mu_0 e^2}{6 m_e} \sum \langle r^2 \rangle

where ∑⟨r2⟩\sum \langle r^2 \rangle is the sum of mean-square orbital radii of all electrons in the atom/molecule.

For a diatomic N₂ molecule, each nitrogen atom has 7 electrons, so 14 electrons total. But the electrons are tightly bound in small orbitals (first-row element). A typical ⟨r2⟩\langle r^2 \rangle for a 2p electron in N is about (0.5 A˚)2=0.25×10−20 m2(0.5 \,\text{Å})^2 = 0.25 \times 10^{-20} \,\text{m}^2. Summing over all electrons gives roughly:

∑⟨r2⟩N2∼14×0.25×10−20≈3.5×10−20 m2\sum \langle r^2 \rangle_{\text{N}_2} \sim 14 \times 0.25 \times 10^{-20} \approx 3.5 \times 10^{-20} \,\text{m}^2

For copper (atomic number 29), the inner electrons (up to 3d) have smaller radii, but the outer 4s electron and especially the 3d electrons have larger orbits. A typical ⟨r2⟩\langle r^2 \rangle for a 3d electron in Cu is about (1.0 A˚)2=1.0×10−20 m2(1.0 \,\text{Å})^2 = 1.0 \times 10^{-20} \,\text{m}^2, and there are 10 such d-electrons. The 4s electron has an even larger orbit, but it contributes less because it’s only one electron. A rough sum:

∑⟨r2⟩Cu∼(core electrons: small)+10×1.0×10−20≈1.0×10−19 m2\sum \langle r^2 \rangle_{\text{Cu}} \sim ( \text{core electrons: small} ) + 10 \times 1.0 \times 10^{-20} \approx 1.0 \times 10^{-19} \,\text{m}^2

That’s about 3 times larger than for N₂. …

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