Rutherford's Experiment that led to Nuclear Model of Atom/Strength of Force

Rutherford's Experiment that led to Nuclear Model of Atom: Strength of Force

Consider Ernest Rutherford's experiment that led him to propose the nuclear model of the atom: that most of its mass is concentrated in one place.

The $\alpha$ particles that were deflected through $180 \degrees$ were coming within $10^{-14} \ \mathrm m$ of the nucleus.

The electric charge on the $\alpha$ particle is $2 \E$, where $\E$ is the elementary charge.

The electric charge on the nucleus of gold is $79 \E$.

Hence by Coulomb's Law of Electrostatics, the force $F$ on the $\alpha$ particle at this range is given by:

\(\ds F\)

\(=\)

\(\ds \dfrac {2 \times 79 \times \E^2} {4 \pi \varepsilon_0 \paren {2 \times 10^{-14} }^2}\)

\(\ds \)

\(\approx\)

\(\ds 90 \ \mathrm N\)

given the values:

$\E = 1.60217 \, 6634 \times 10^{−19} \ \mathrm C$

$\varepsilon_0 = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$

Hence we have a force acting within a single atom of the weight of nearly $10 \mathrm {kg}$.

$\blacksquare$

Sources

• 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge