An alpha particle's trajectory is deflected by 60° in a Rutherford scattering experiment. What is the impact parameter (b) for this event, expressed in terms of the distance of closest approach (

Question

An alpha particle's trajectory is deflected by 60° in a Rutherford scattering experiment. What is the impact parameter (b) for this event, expressed in terms of the distance of closest approach ( r 0 ​ ) for a head-on collision at the same kinetic energy?

TestHub · Accepted Answer

Distance of closest approach and impact parameter: $r_{0} = \frac{kZe ^{2}}{KE}$ and $b = \frac{k ( Ze ^{2} ) c o t ( θ /2 )}{2 KE}$

Substituting $r_{0}$ : $b = r_{0} \frac{c o t ( θ /2 )}{2}$

For $θ = 6 0^{\circ}$ : $b = \frac{r _{0}}{2} cot (3 0^{\circ}) = \frac{r _{0}}{2} \cdot 3 = \frac{3}{2} r_{0}$ hence (B).

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