The kinetic energy of an \( \alpha \)-particle incident on gold foil is reduced to half of its initial value. How does the distance of closest approach change?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → becomes twice
At the distance of closest approach \( r_0 \) the α-particle stops for an instant, so all its initial kinetic energy K has been converted into electrostatic potential energy of the α-particle (charge \( 2e \)) and the gold nucleus (charge \( Ze \)).
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0} \)
\( \displaystyle r_0 = \frac{2Ze^{2}}{4\pi\varepsilon_0 K} \propto \frac{1}{K} \)
When K becomes \( K/2 \), the distance of closest approach becomes \( 2r_0 \).