An alpha particle colliding with one of the electrons in a gold atom loses

An alpha particle is projected towards a stationary ${}_{92}^{235}U$ nucleus with $KE$ kinetic energy. Find the distance of closest approach.

An $\alpha$-particle of energy $4 \text{ MeV}$ is scattered through $180^\circ$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

$A$ proton is fired from very far away towards a nucleus with charge $Q=120 \ e$,where $e$ is the electronic charge. It makes a closest approach of $10 \ fm$ to the nucleus. The de Broglie wavelength (in units of $fm$) of the proton at its start is: (take the proton mass,$m_0 = (5/3) \times 10^{-27} \ kg$,$h/e = 4.2 \times 10^{-15} \ J \cdot s/C$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$,$1 \ fm = 10^{-15} \ m$)

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

In $1911$,the physicist Ernest Rutherford discovered that atoms have a tiny,dense nucleus by shooting positively charged particles at a very thin gold foil. $A$ key physical property which led Rutherford to use gold was that it was