Amit & Deepali Sinha Foundation

One Free Electron per Sodium Atom

Irodov 3.267DifficultyUniqueness

Problem Irodov 3.267

A Hall effect measurement is made on a sodium conductor (density 0.970.97 g/cm3^3, molar mass 2323 g/mol). With a current density j=200j = 200 A/cm2^2 in a magnetic field of induction B=1.00B = 1.00 T, the transverse electric field is found to be E=5.0E = 5.0 μ\muV/cm. Find the concentration of conduction electrons, and its ratio to the concentration of atoms in the sodium.

The sodium sample carrying the current density j across the field B (its sense is not given; drawn into the page).

The sodium sample carrying the current density j\vect j across the field B\vect B (its sense is not given; drawn into the page).

Given
  • E=5.0E = 5.0 μ\muV/cm
  • j=200j = 200 A/cm2^2
  • B=1.00B = 1.00 T
  • sodium, density 0.970.97 g/cm3^3 (table value)
  • molar mass 2323 g/mol (table value)
  • j⊥B\vect j \perp \vect B (assumed)
Find
the electron concentration nn and its ratio to the atom concentration

Solution

Step 1The Hall field balances the magnetic force
The steady state. The electrons drift with velocity u against j; the magnetic force F_m drives them to the upper face, which charges negative, and the transverse field E of the separated charges points from the positive face to the negative one.

The steady state. The electrons drift with velocity u\vect u against j\vect j; the magnetic force Fm\vect F_m drives them to the upper face, which charges negative, and the transverse field E\vect E of the separated charges points from the positive face to the negative one.

The carriers are electrons, of charge −e-e, drifting against the current with the speed u=j/neu = j/ne, at right angles to B\vect B (assumed). The magnetic force −e u×B-e\,\vect u\times\vect B pushes them across the current, and they pile up on one face until the transverse field E\vect E of the separated charge stops further sideways drift:

eE=euB=e jne BeE = euB = e\,\frac{j}{ne}\,B
E=jBneE = \frac{jB}{ne}
(1)
Step 2The electron concentration

Solving (1) for nn gives the form in the box below. In SI units j=2.00×106j = 2.00\times 10^6 A/m2^2 and E=5.0×10−4E = 5.0\times 10^{-4} V/m, so numerically

n=2.00×106×1.001.60×10−19×5.0×10−4 m−3n = \frac{2.00\times 10^6\times 1.00}{1.60\times 10^{-19}\times 5.0\times 10^{-4}}\ \text{m}^{-3}
n=jBeE≈2.5×1028 m−3n = \frac{jB}{eE} \approx 2.5\times 10^{28}\ \text{m}^{-3}
(2)
Step 3The concentration of atoms

Sodium’s density ρ=0.97×103\rho = 0.97\times 10^3 kg/m3^3 and molar mass M=23×10−3M = 23\times 10^{-3} kg/mol are table values. A cubic metre holds ρ/M\rho/M moles, so the concentration of atoms is

na=ρNAM=0.97×103×6.02×102323×10−3≈2.54×1028 m−3(3) n_a = \frac{\rho N_A}{M} = \frac{0.97\times 10^3\times 6.02\times 10^{23}}{23\times 10^{-3}} \approx 2.54\times 10^{28}\ \text{m}^{-3} \tag{3}

Dividing (2) by (3):

nna=2.5×10282.54×1028\frac{n}{n_a} = \frac{2.5\times 10^{28}}{2.54\times 10^{28}}
nna≈0.98\frac{n}{n_a} \approx 0.98
(4)

There is almost exactly one conduction electron per sodium atom.

Check

Units: jB/eEjB/eE is (A/m2^2)(V⋅\cdots/m2^2)/(C⋅\cdotV/m) == (C/s)(s)/(C⋅\cdotm3^3) == m−3^{-3}, a concentration. Chemistry agrees: sodium is an alkali metal with a single valence electron outside a closed shell, so one free electron per atom is expected, and the ratio comes out as 11 within the two-figure precision of the data.

Remark

The Hall field is tiny, a few microvolts per centimetre even at a tesla and a large current density, because E=jB/neE = jB/ne falls as the carrier concentration rises; in a semiconductor, with 10610^{6} times fewer carriers, the same measurement gives volts per centimetre. The measurement counts carriers directly, without any model of how they move, and its sign tells whether they are electrons or holes.