Question

A proton of energy 1.6 MeV approaches a gold nucleus (Z = 79). Find the distance of its closest approach.

Answer 1

To find the distance of closest approach of a proton to a gold nucleus, we use the principle of energy conservation. At the closest point, all the kinetic energy of the proton is converted into electrostatic potential energy due to repulsion from the positively charged gold nucleus.

Formula for Distance of Closest Approach

`d = 1/(4pi∈_0) xx (Ze^2)/E`

Where:

d = distance of closest approach (in meters)

Z = 79 (atomic number of gold)

e = 1.6 × 10⁻¹⁹C (charge of proton)

E = 1.6 MeV = 1.6 × 10⁶ × 1.6 × 10⁻¹⁹ J = 2.56 × 10⁻¹³J

`1/(4pi∈_0) = 9 xx 10^9` Nm²/C²

`d = 9xx10^9xx(79xx(1.6xx10^-19)^2)/2.56xx10^-13`

79 ⋅ (1.6 × 10⁻¹⁹)² = 79 ⋅ 2.56 × 10⁻³⁸ = 2.0224 × 10⁻³⁶

`d = (9xx10^9xx2.0224xx10^-36)/(2.56xx10^-13)`

`(1.8202xx10^-26)/(2.56xx10^-13)`

d = 7.11 × 10⁻¹⁴m