In an atom,for the electron to revolve around the nucleus,the necessary centripetal force is obtained from which of the following forces exerted by the nucleus on the electron?

$\alpha$-particles of energy $400 \, KeV$ are bombarded on a nucleus of $_{82}Pb$. In the scattering of $\alpha$-particles, what will be the minimum distance of approach from the nucleus?

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

When an $\alpha$-particle of mass $m$ moving with velocity $v$ bombards a heavy nucleus of charge $Ze$,its distance of closest approach from the nucleus depends on $m$ as

How is an atom represented? Why?

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 $1.5 P$?