Geiger-Marsden alpha particle scattering — Rutherford's atomic model

Question

Describe the Geiger-Marsden alpha particle scattering experiment. What were the key observations? How did Rutherford use these results to propose the nuclear model of the atom? Calculate the distance of closest approach for a 5 MeV alpha particle hitting a gold nucleus.

(NCERT Class 12, Atoms)

Solution — Step by Step

Geiger and Marsden (1911), under Rutherford’s supervision, directed a beam of alpha particles ( $\alpha$ , which are He²⁺ nuclei with charge +2e) at a thin gold foil (about 1000 atoms thick). A movable zinc sulphide (ZnS) screen surrounding the foil detected scattered alpha particles by producing scintillations (flashes of light).

Most alpha particles (over 99%) passed straight through the foil with little or no deflection → most of the atom is empty space
A small fraction were deflected by moderate angles → some positive charge present to repel alpha particles
About 1 in 8000 alpha particles bounced back at angles greater than 90° (some nearly 180°) → the positive charge and most of the mass are concentrated in a very small, dense region

From these observations, Rutherford proposed:

The atom has a tiny, dense, positively charged centre called the nucleus (radius ~ $10^{-15}$ m)
The nucleus contains almost all the mass of the atom
Electrons orbit the nucleus at relatively large distances (atomic radius ~ $10^{-10}$ m)
The atom is mostly empty space (ratio of nuclear to atomic radius ~ $10^{-5}$ )

The backward scattering could only be explained if the entire positive charge was concentrated in a region much smaller than the atom — a diffuse positive charge (as in Thomson’s model) could never produce such large deflections.

For a head-on collision (180° scattering), all kinetic energy converts to electrostatic potential energy at the distance of closest approach $d$ :

\frac{1}{2}m_\alpha v^2 = \frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{d}

d = \frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{KE}

For gold ( $Z = 79$ ) and $KE = 5$ MeV $= 5 \times 1.6 \times 10^{-13}$ J:

d = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{5 \times 1.6 \times 10^{-13}}

= \frac{9 \times 10^9 \times 158 \times 2.56 \times 10^{-38}}{8 \times 10^{-13}}

= \frac{3.64 \times 10^{-26}}{8 \times 10^{-13}} = \mathbf{4.55 \times 10^{-14} \text{ m}} \approx 45.5 \text{ fm}

This gives an upper bound for the nuclear radius of gold.

Why This Works

The alpha particle scattering experiment is a beautiful example of using projectiles as probes. Fast-moving, positively charged alpha particles interact with the atom’s positive charge via the Coulomb force. The scattering pattern reveals the charge distribution.

Thomson’s “plum pudding” model (positive charge spread uniformly over the atom) would produce only small-angle deflections — the distributed charge can’t create a strong enough electric field to turn an alpha particle around. Only a concentrated point-like charge (the nucleus) can produce the enormous electric field needed for large-angle and backward scattering.

The distance of closest approach calculation uses energy conservation: at the turning point, the alpha particle momentarily stops, and all its kinetic energy has been converted to Coulomb potential energy. This distance is slightly larger than the actual nuclear radius.

Alternative Method

For quick estimation of the distance of closest approach, use the formula in convenient units:

d = \frac{kZe \times 2e}{KE} = \frac{1.44 \times 2Z}{KE(\text{MeV})} \text{ fm}

where $ke^2 = 1.44$ MeV·fm is a useful constant. For gold with 5 MeV:

d = \frac{1.44 \times 2 \times 79}{5} = \frac{227.5}{5} = 45.5 \text{ fm}

JEE Main frequently asks for the distance of closest approach or the ratio of distances for different energies/nuclei. Remember: $d \propto Z/KE$ . If you double the alpha particle energy, $d$ halves. If you use a nucleus with double the atomic number, $d$ doubles. These proportionality questions are quick 2-minute problems.

Common Mistake

Students often write that Rutherford’s experiment “proved atoms have electrons orbiting the nucleus.” The experiment actually said nothing about electrons — it only revealed the existence of a small, dense, positive nucleus. The idea of electrons orbiting came from the need to balance the positive nuclear charge with negative charges somewhere in the atom. Rutherford assumed circular orbits, but this model had its own problems (orbiting electrons should radiate energy and spiral inward), which Bohr later resolved.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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