Question

The ‘distance of closest approach’ of an alpha-particle is \(d\) when it moves with a velocity \(v\) head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –

• A. \(\dfrac{d}{2}\)
• B. \(2d\)
• C. \(\dfrac{d}{4}\)
• D. \(4d\)

Hint:

For a head-on collision of an alpha particle with a nucleus, \[ d\propto \frac{1}{v^2}. \] If the velocity becomes \(n\) times smaller, the distance of closest approach becomes \(n^2\) times larger.

Answer 1

The Correct Option is C

Concept: The distance of closest approach is the minimum distance between the alpha particle and the nucleus during a head-on collision. At this point, the entire initial kinetic energy of the alpha particle gets converted into electrostatic potential energy. According to the principle of conservation of energy, \[ \text{Initial K.E.}=\text{Electrostatic P.E. at closest approach} \] For an alpha particle of charge \(+2e\) approaching a nucleus of charge \(+Ze\), \[ \frac{1}{2}mv^2=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r} \] where \(r\) is the distance of closest approach. From this expression, it is evident that \[ r\propto \frac{1}{v^2}. \] Thus, the distance of closest approach is inversely proportional to the square of the speed of the alpha particle.

Step 1: Write the proportionality relation.
\[ d\propto \frac{1}{v^2} \] Suppose the new distance of closest approach is \(d'\) when the velocity becomes \[ v'=\frac{v}{2}. \]

Step 2: Form the ratio of the two distances.
\[ \frac{d'}{d} = \frac{\dfrac{1}{(v/2)^2}}{\dfrac{1}{v^2}} \] \[ = \frac{v^2}{v^2/4} \] \[ =4 \] Hence, \[ d'=4d. \] \[ \boxed{d'=4d} \] Therefore, the new distance of closest approach becomes four times the original value.