The size of an atom is of the order of:

Show the trajectory of $\alpha$-particle for different impact parameters and explain how Rutherford used this to determine the upper limit of the nuclear size.

An $\alpha$-particle of $5 \; MeV$ energy strikes a stationary uranium nucleus at a scattering angle of $180^o$. The distance of closest approach of the $\alpha$-particle to the nucleus will be 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,$m_0 = (5/3) \times 10^{-27} \ kg$,$h/e = 4.2 \times 10^{-15} \ J \cdot s/C$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$,$1 \ fm = 10^{-15} \ m$)

$A$ fundamental particle of mass $m$ and charge $+e$ is projected towards a much heavier charged particle of charge $+Ze$,where $Z > 0$. The distance of closest approach of the incident particle is ........

An $\alpha$ particle is accelerated through a potential difference of $10^6 \ V$. Then the kinetic energy $(K.E.)$ of the particle will be .......... $MeV$.