Question

In Rutherford scattering experiment, the correct angle for scattering of \[\alpha - \]particles for impact parameter equal to zero is
A. \[90^\circ \]
B. \[120^\circ \]
C. \[270^\circ \]
D. \[180^\circ \]

Answer 1

Hint: The impact parameter is the distance between the path of \[\alpha - \]particle and the field of the object placed at the centre. The angle from which the \[\alpha - \]particles are scattered from the horizontal path is known as angle of scattering. Use the relation between scattering angle and impact parameter to measure the angle of scattering.

Complete step by step answer:
To answer this question, we need to understand the impact parameter and scattering angle. The impact parameter is the distance between the path of \[\alpha - \]particle and the field of the object placed at the centre. Since the \[\alpha - \]particles are negatively charged, the path of the particles is deflected from the field created by the positive charges in the nucleus of gold foil. The angle from which the \[\alpha - \]particles are scattered from the horizontal path is known as angle of scattering.

We know that the impact parameter is proportional to \[\cot \dfrac{\varphi }{2}\].
\[\therefore b \propto \cot \dfrac{\varphi }{2}\], where, \[\varphi \] is the angle of scattering.

Since we have given that the impact parameter is zero, the above equation will become,
\[0 = \cot \dfrac{\varphi }{2}\]
\[ \Rightarrow \varphi = 2{\cot ^{ - 1}}\left( 0 \right)\]
\[ \Rightarrow \varphi = 2\left( {90^\circ } \right)\]
\[ \Rightarrow \varphi = 180^\circ \]

Therefore, the angle of scattering is \[180^\circ \] for impact parameter 0. Since the angle of scattering is \[180^\circ \], the alpha particles will bounce back into their incident path after deflection from the nucleus.

Note:You don’t need to solve the expression for scattering angle for deflection of \[\alpha - \] particles if you know the relation between the scattering angle and impact parameter. One can answer this question, if the definition of impact parameter is known. Since the impact parameter is zero, the \[\alpha - \] particle is directly incident on the nucleus. We know that the nucleus has positive charge, \[\alpha - \] particles will surely deflect back in their original path.