Question

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

Answer 1

The Rutherford scattering formula can be used to calculate the closest approach distance between a proton and a gold nucleus. Rutherford scattering describes the deflection of charged particles, such as alpha particles or protons, when they enter a nucleus. When an incident particle with charge q and kinetic energy K approaches a nucleus with charge Ze and radius R, the Rutherford scattering formula for the distance of the closest approach (r_min) is as follows:

`r_(min) = 1/(4piepsilon_0) (2Ze^2)/K`

Where:

• K is the kinetic energy of the incident particle (in joules),
• Z is the atomic number of the nucleus (number of protons), and
• e is the elementary charge (1.6 × 10⁻¹⁹ C).

Given that the energy of the proton is 1.6 MeV,

∵ 1 MeV = 1.6 × 10⁻¹³ J

∴ 1.6 MeV = 1.6 × 1.6 × 10⁻¹³ J

= 2.56 × 10⁻¹³ J

Now,

`r_(min) = 9.0 xx 10^9 xx (2 xx 79 xx (1.6 xx 10^-19))/(2.56 xx 10^-13)`

= 1422 × 10⁻¹⁶ m

r_min ≈ 1.42 × 10⁻¹³ m

So, the distance of closest approach is approximately 1.42 × 10⁻¹³ m.