Question

An alpha particle of energy $5 \,MeV$ is scattered through $180^\circ$ by a fixed uranium nucleus. The distance of closest approach is of the order of

• A. $1\, \mathring{A}$
• B. $10^{-10} cm$
• C. $10^{-12} cm$
• D. $10^{-15} cm$

Answer 1

The Correct Option is C

From conservation of mechanical energy decrease in kinetic energy = increase in potential energy or \(\frac{1}{4 \pi \varepsilon_0} \frac{(Ze)(Ze)}{r_{min}}=5\, MeV=5 \times 1.6 \times10^{-13}\, J\)

\(\therefore \, \, \, r_{min}=\frac{1}{4 \pi \varepsilon_0}\frac{2Ze}{5 \times 1.6 \times10^{-13}}\)

\(15mm = \frac{(9 \times 10^9) (2)(92)(1.6 \times 10^{-19})^2}{5 \times 1.6 \times10^{-13}}\)

\(15mm = 5.3 \times 10^{-14}\, m= 5.3 \times 10^{-12} cm\) 

i.e.\(r_{min}\) is of the order of \(10^{-12} cm\)

Answer 2

\(\text{K.E.} = \left( \frac{1}{2} \right) \left( \frac{1}{4\pi\epsilon_0} \right) \frac{Z e^2 e}{d} = 5 \times 10^6 \times 1.6 \times 10^{-19} = 9 \times 10^9 \times 235 \times 2 \times \left( \frac{1.6 \times 10^{-19}}{2} \right)^2 d\)

\(d = \frac{9 \times 10^9 \times 470 \times 1.6 \times 10^{-19}}{5 \times 10^6} \, \text{meters}\)

\(𝑑=10^{−12} cm\)