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,$m_0 = (5/3) \times 10^{-27} \ kg$,$h/e = 4.2 \times 10^{-15} \ J \cdot s/C$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$,$1 \ fm = 10^{-15} \ m$)

According to classical theory,the circular path of an electron in a Rutherford atom is

Explain the Rutherford atomic model and its limitations.

Based on which experiment did the Rutherford nuclear model come from?

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?