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?

$(a)$ about the same

The average angle of deflection of $\alpha$ -particles by a thin gold foil predicted by Thomson's model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models.

$(b)$ Much less

The probability of scattering of $\alpha$ -particles at angles greater than $90^{\circ}$ predicted by Thomson's model is much less than that predicted by Rutherford's model.

$(c)$ Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

$(d)$ Thomson's model

It is wrong to ignore multiple scattering in Thomson's model for the calculation of average angle of scattering of $\alpha$ particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.