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
The initial kinetic energy (K) of the $α$-particle is entirely converted into electrostatic potential energy at distance of closest approach ($r_{min}$):
$K=\frac{1}{4πε_0}\frac{Ze^2}{r_{min}}$
Z, Atomic number of gold nucleus = 79
e, Elementary charge = $1.6×10^{-19}$
$∴r_{min}=\frac{1}{4πε_0}\frac{Ze^2}{K}$
Now,
if the kinetic energy is reduced to half of its initial value ($k→k/2$)
$r_{min}=\frac{1}{4πε_0}\frac{Ze^2}{K/2}$
$⇒r_{min}=2.\frac{1}{4πε_0}\frac{Ze^2}{K}=2.r_{min}$