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The NCERT Class 12 chapter on the dual nature of radiation gives three ways to get electrons out of a metal: heat it, pull them out with a strong electric field, or shine light on it. Most of the chapter is about the third. This post is about the first, thermionic emission, and the physicist who found its law: Owen Richardson.
Who was Owen Richardson?
Richardson was born on 26 April 1879 in Dewsbury, Yorkshire, according to his Nobel Prize biography. He went to Trinity College, Cambridge in 1897, and after graduating in 1900 began to investigate the emission of electricity from hot bodies at the Cavendish Laboratory. He was Professor of Physics at Princeton from 1906 to 1913, was elected a Fellow of the Royal Society in 1913, and in 1914 became Wheatstone Professor of Physics at King's College London. He was knighted in 1939 and died on 15 February 1959.
He received the Nobel Prize in Physics 1928 "for his work on the thermionic phenomenon and especially for the discovery of the law named after him". The prize was awarded one year later, in 1929.
What did Richardson discover?
Physicists already knew that air near a hot wire conducts electricity. Many thought gas around the wire was responsible. In his Nobel Lecture, Richardson describes deciding to remove the gas and see what was left. Pumps were poor, and heating the wire drove gas out of the tube walls, so he sometimes heated a wire in a tube for weeks to be sure the current was not coming from leftover gas. There was no ductile tungsten yet, so he used platinum.
In 1901 he showed that each unit area of a hot platinum surface emitted a limited number of electrons, and that the number rose very rapidly with temperature. His Nobel biography quotes the law as he announced it to the Cambridge Philosophical Society on 25 November 1901: the saturation current should obey s = AT^(1/2)e^(−b/T).
In 1911 he concluded that a form with T² in place of T^(1/2) was theoretically better. Today it is usually written J = AT²e^(−W/kT), where J is current per unit area, W is the work function and k is Boltzmann's constant. In 1913, with ductile tungsten, he drew currents large enough to show that the mass of electrons emitted was more than the mass of any chemicals that could have been used up. He wrote that this experiment ended the argument that the effect was just a chemical reaction.
A worked example: how much does a little heat matter?
Thermionic emission and Richardson's law are not listed in the JEE (Main) 2026 syllabus, but this example uses only exponentials and the electron volt from Class 12.
Question (sample numbers). A filament has work function W = 4.5 eV. By what factor does the emission current rise when it is heated from 2000 K to 2500 K? Use k = 8.617 × 10⁻⁵ eV/K, the NIST value.
Write the ratio: J₂/J₁ = (T₂/T₁)² × e^[(W/k)(1/T₁ − 1/T₂)].
W/k = 4.5 ÷ (8.617 × 10⁻⁵) ≈ 52,200 K.
1/2000 − 1/2500 = 0.0005 − 0.0004 = 0.0001 per kelvin.
Exponent: 52,200 × 0.0001 = 5.22, and e^5.22 ≈ 185.
T² factor: (2500/2000)² = 1.25² = 1.5625.
Ratio ≈ 1.5625 × 185 ≈ 290.
A 25% rise in temperature gives about 290 times the current. Almost all of it comes from the exponential. If you use Richardson's 1901 T^(1/2) form instead, the prefactor is about 1.118 and the ratio is about 207, still the same order of size. That is the point Richardson made in his Nobel Lecture: it is impossible to tell the two forms apart by experiment, because the effect of the T² or T^(1/2) term is so small next to the exponential.
Common mistakes students make
Using degrees Celsius. T must be in kelvin, both in T² and in the exponent.
Mixing eV and joules. 1 eV = 1.602 × 10⁻¹⁹ J, as NCERT's Dual Nature chapter gives. Convert once, at the start.
Expecting a linear rise. Doubling the temperature does not double the current; the exponential makes it rise far faster.
Confusing thermionic and photoelectric emission. Both need electrons to overcome the work function. Heat supplies the energy in one, a photon in the other. Richardson's lecture notes the link: the photoelectric threshold frequency ν₀ and work function w₀ are related by w₀ = hν₀.
For the photon side of the story, see our guide to photoelectric effect problems. If you get the exponent wrong and cannot see why, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
What can a student take from Richardson's way of working?
Remove the complications first. He thought gas was hiding the real effect, so he got rid of the gas, even when that took weeks.
Let better tools settle old arguments. When ductile tungsten arrived in 1913, he used it to run the decisive experiment.
Know what your data can and cannot tell you. He admitted that experiments could not choose between T² and T^(1/2), and chose on theoretical grounds.
Redo the worked example with T₂ = 2200 K, then check your answer by working out the exponent and the T² factor separately.
