Question

The distance-of-closest-approach for an alpha particle is ‘d’ when it moves head-on with speed v towards a target nucleus. If the alpha particle is replaced by a proton moving with the same speed, the new distance-of-closest-approach will be ______.

पर्याय

• 2d

• d

• `d/2`

• `d/sqrt2`

Answer 1

The distance-of-closest-approach for an alpha particle is ‘d’ when it moves head-on with speed v towards a target nucleus. If the alpha particle is replaced by a proton moving with the same speed, the new distance-of-closest-approach will be `bbunderline(d/2)`.

Explanation:

The distance of closest approach is found using conservation of energy:

`1/2 mv^2 = 1/(4 pi epsilon_0) (Z ze^2)/r`

So,

r ∝ `(Z z)/(m v^2)`

For the same target and the same speed,

r ∝ `z/m`

For alpha particle properties:

Charge = 2e

Mass ≈ 4m_p

So, r_α ∝ `2/4 = 1/2`

For proton properties:

Charge = e

Mass = m_p

So, r_p ∝ `1/1` = 1

∴ `r_p/r_alpha = 1/(1//2)`

= 2

So, the proton goes twice as far compared to the alpha particle reference scaling.

Given alpha distance = d corresponds to a proportional factor of 1/2. Thus, proton distance corresponds to factor 1 → half relative scaling:

r_p = `d/2`