Question

In alpha particle scattering experiment, if v is the initial velocity of the particle, then the distance of closest approach is d. If the velocity is doubled, then the distance of closest approach becomes

• A. 4 d
• B. 2 d
• C. $\frac{d}{2}$
• D. $\frac{d}{4}$

Answer 1

The Correct Option is D

In alpha particle scattering, the closest approach $d$ is inversely proportional to the square of the initial velocity of the particle. This is derived from the energy conservation principle in the scattering experiment. The equation can be expressed as: $$d \propto \frac{1}{v^2}$$ This means if the initial velocity $v$ is doubled, the distance of closest approach will change by a factor of $\frac{1}{4}$, because: $$d' = \frac{d}{(2)^2} = \frac{d}{4}$$ Thus, the new closest approach is $\frac{d}{4}$. Thus, the solution is $\frac{d}{4}$.