Question
In Rutherford scattering experiment, what will be the correct angle for $$\alpha $$-scattering for an impact parameter, $$b = 0$$ ?
A.
$${90^ \circ }$$
B.
$${270^ \circ }$$
C.
$${0^ \circ }$$
D.
$${180^ \circ }$$
Answer :
$${180^ \circ }$$
Solution :
Impact parameter is perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
Rutherford calculated analytically, the relation between the impact parameter $$b$$ and scattering angle $$\theta ,$$ which is given by
$$b = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Z{e^2}\cot \frac{\theta }{2}}}{E}$$
where, $$E = \frac{1}{2}m{v^2}$$ is kinetic energy of alpha particle, when it is far away from the atom. According to problem,
$$b = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Z{e^2}\cot \frac{\theta }{2}}}{E} = 0$$
As given that $$b = 0$$
$$\eqalign{
& {\text{so,}}\,\,\cot \frac{\theta }{2} = 0 \cr
& \therefore \frac{\theta }{2} = {90^ \circ }\,\,{\text{or}}\,\,\theta = {180^ \circ } \cr} $$