Take the energy barrier for thermal reversal of the magnetization of a single-domain particle of volume V as KV (K is the magnetic anisotropy constant); for a measurement time of about 100 s, KV = 25kT is a guide to the boundary for superparamagnetism. With K = 1.1×10⁴ J/m³, a temperature of 300 K and the Boltzmann constant k = 1.38×10⁻²³ J/K, which value is closest to the upper limit of the diameter of a spherical particle that becomes superparamagnetic?

Answer: C. About 26 nm

The boundary volume is V = 25kT/K = 25 × 1.38×10⁻²³ × 300/(1.1×10⁴) ≈ 9.4×10⁻²⁴ m³. From the volume of a sphere V = πd³/6, d = (6V/π)^(1/3) ≈ 2.6×10⁻⁸ m, so particles with diameters of about 26 nm or less become superparamagnetic. About 13 nm is the radius, given as if it were the diameter, and about 9 nm drops the factor of 25 and uses KV = kT. The factor of 25 comes from ln(100 s/10⁻⁹ s) ≈ 25, taking the attempt time for reversal as about 10⁻⁹ s. The diameter is inversely proportional to the cube root of K, so materials with larger anisotropy can keep their magnetization stable even in smaller particles.