An $\alpha$-particle of energy $5 \text{ MeV}$ is scattered through $180^{\circ}$ by a gold nucleus. The distance of closest approach is of the order of:

Answer the following questions,which help you understand the difference between Thomson's model and Rutherford's model better.
$(a)$ Is the average angle of deflection of $\alpha$-particles by a thin gold foil predicted by Thomson's model much less,about the same,or much greater than that predicted by Rutherford's model?
$(b)$ Is the probability of backward scattering (i.e.,scattering of $\alpha$-particles at angles greater than $90^{\circ}$) predicted by Thomson's model much less,about the same,or much greater than that predicted by Rutherford's model?
$(c)$ Keeping other factors fixed,it is found experimentally that for small thickness $t$,the number of $\alpha$-particles scattered at moderate angles is proportional to $t$. What clue does this linear dependence on $t$ provide?
$(d)$ In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of $\alpha$-particles by a thin foil?

What radioactive source did Geiger and Marsden use in the scattering experiment?

If an alpha particle with energy $7.7 \text{ MeV}$ is bombarded on a thin gold foil, the closest distance from the nucleus it can reach is . . . . . . $\text{m}$. (Atomic number of gold $= 79$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ in $SI$ units)

In a Geiger-Marsden experiment,what is the distance of closest approach to the nucleus of a $7.7 \;MeV$ $\alpha$-particle before it comes momentarily to rest and reverses its direction?

An $\alpha$-particle of energy $5 \, MeV$ is scattered by a fixed uranium nucleus at $180^o$. What is the distance of closest approach between the particle and the uranium nucleus?