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 \( \frac{d}{4} \). 

Answer 2

In the alpha particle scattering experiment, the distance of closest approach \( d \) is related to the initial velocity \( v \) of the particle. The formula for the distance of closest approach is given by: \[ d \propto \dfrac{1}{v^2} \] This relation implies that the distance of closest approach is inversely proportional to the square of the velocity. If the velocity is doubled, the new velocity \( v' = 2v \). Now, using the proportionality: \[ d' \propto \dfrac{1}{(2v)^2} = \dfrac{1}{4v^2} \] Thus, the new distance of closest approach is: \[ d' = \dfrac{d}{4} \] Therefore, when the velocity is doubled, the distance of closest approach becomes \( \dfrac{d}{4} \).