If in Rutherford's experiment,the number of particles scattered at $90^o$ angle are $28$ per min,then the number of scattered particles at an angle $60^o$ and $120^o$ will be:

Which scientist proposed the plum pudding model of the atom?

$\sqrt{d_{1}}$ and $\sqrt{d_{2}}$ are the impact parameters corresponding to scattering angles $60^{\circ}$ and $90^{\circ}$ respectively,when an $\alpha$-particle is approaching a gold nucleus. For $d_{1} = x d_{2}$,the value of $x$ will be . . . . . .

In an alpha particle scattering experiment,the distance of closest approach for the $\alpha$-particle is $4.5 \times 10^{-14} \ m$. If the target nucleus has an atomic number $Z = 80$,then the maximum velocity of the $\alpha$-particle is approximately $... \times 10^5 \ m/s$.
$\left(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ SI \ unit, \text{mass of } \alpha \text{-particle } m = 6.72 \times 10^{-27} \ kg, e = 1.6 \times 10^{-19} \ C\right)$

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:

In Rutherford's experiment,the number of particles scattered at a $90^{\circ}$ angle is $x$ per second. The number of particles scattered per second at an angle of $60^{\circ}$ is: