Skip to content

Chemistry · Ch 4 — Structure of Atom

Discovery of Proton

4.1.2

Discovery of Proton

In 1911, Ernest Rutherford investigated the structure of the atom by firing a narrow beam of positively-charged α-particles at an extremely thin sheet of gold foil, with a fluorescent screen arranged around the foil to record where each particle landed after passing through. Most α-particles passed straight through the foil as if nothing were there, a smaller fraction were deflected through small angles, and — the genuinely surprising result — a very small number bounced almost straight back the way they had come. Rutherford reasoned that such large deflections could only happen if the α-particles were occasionally colliding with something extremely small, extremely dense, and positively charged (since like charges repel); everything else about the result implied that atoms are mostly empty space. This experiment established that an atom has a tiny, massive, positively-charged nucleus at its centre. Rutherford followed this up in 1919 by showing that bombarding nitrogen gas with fast α-particles converts it into oxygen while simultaneously releasing hydrogen nuclei: 714N+24α→ 817O+11H^{14}_{7}N + ^{4}_{2}\alpha \rightarrow\, ^{17}_{8}O + ^{1}_{1}H. Because this hydrogen nucleus turned up as a product of transmuting several different elements, Rutherford proposed that every atomic nucleus contains hydrogen nuclei as one of its building blocks, …

Figure 4.2Rutherford's scattering experiment

What this figure shows. A diagram of the α-particle scattering set-up: a radioactive source emits a narrow beam of α-particles toward a very thin sheet of gold foil, with a circular fluorescent (zinc-sulphide) detector screen surrounding the foil to record where the α-particles land after passing the foil. The diagram shows most particles travelling straight through the foil undeflected, a smaller number deflected through small angles, and a very small number bouncing almost straight back toward the source — the observation from which Rutherford inferred that an atom's positive charge and most of its mass are concentrated in a tiny, den …

Misc Problem 4.1Composition of the ¹⁸⁴⁰Ar nuclide

Worked out. Worked example: find the number of protons, electrons and neutrons in the nuclide 1840Ar^{40}_{18}Ar. For this nuclide the mass number A = 40 and the atomic number Z = 18. Since number of protons = number of electrons = Z, both equal 18. The number of neutrons N = A − Z = 40 − 18 = 22. So 1840Ar^{40}_{18}Ar has 18 protons, 18 electrons and 2 …