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 ______.

Options

• 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 that an alpha distance of d corresponds to a proportional factor of `1/2`, a proton distance corresponds to a factor of 1. This represents a half-relative scaling.

r_p = `d/2`