An alpha particle with kinetic energy 20 MeV is heading towards a stationary tin nucleus of atomic number 50, Compute the distance of closest approach?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)

Answer 1

Concept:

The closest distance of approach of the α particle will be when the whole kinetic energy of the α particle will be converted into potential energy.

\(K.E. = \frac{1}{4π ϵ _o}× \frac{2Ze^2}{r}\)

Where, r = Closest distance of approach, e = Charge of a electron, Z = Atomic number, K.E. = 20 × 1.6 × 10^-13J

Calculation:

Given: e = 1.6 × 10^-19Coulombs, K.E. = 32 × 10^-13jJ, Z = 50 and \(\frac{1}{4\pi \epsilon_o} = 9\times 10^9\)

_{\(32\times 10^{-13} = 9\times 10^9\times \frac{2\times 50 \times (1.6\times 10^{-19})^2}{r} \)}

_{\(\Rightarrow r = \frac{2.304\times 10^{-26}}{32\times 10^{-13}}\)}

\(\Rightarrow r =7.2\times 10^{-15}m\)

r = 7.2 fm