Question

Find out the area of the gold foil used in Rutherford's experiment ?

Answer 1

In Rutherford's gold foil experiment, alpha particles deflect when they collide with the thin gold foil.
So, by applying inverse-square law between the charges on the alpha particles and nucleus, we can write,

$\frac{1}{2}m{v}^2=\frac{1}{4\mathrm{\pi }{\in }_0}\times \frac{q_1{q}_2}{b}$

m(mass) = 6.64424 x 10^-27 kg = 3.7273 x 10⁹ eV
q₁ = 2 x (1.6 x 10^-19) C
​q₂ (for gold) = 79 x (1.6 x 10^-19) C
v (initial velocity) = 2 x 10⁷

Substituting the values, we get, b = 2.7 x 10^-14m

Moreover this gold foil will be approximately of 8.6 x 10^-6 cm.