Question

In Rutherford experiment, a $5.3MeV$ alpha particle moves towards the gold nucleus $(Z=79)$ . How close does the alpha particle get to the centre of the nucleus, before it comes momentarily to rest and reverses its motion?
$({\epsilon }_0=8.8\times 10^{-12}F/m)$

• A. $3.4\times 10^{-15}m$
• B. $8.6\times 10^{-14}m$
• C. $4.3\times 10^{-14}m$
• D. $1.6\times 10^{-14}m$

Answer 1

Answer: (C)

The kinetic energy of $\alpha$ - particle is completely converted into potential energy
$\frac{1}{2}m{u}^2=\frac{1}{4\pi {\epsilon }_0}\frac{(ze)(ze)}{r_0}$
$\therefore$ Distance of closest approach
$r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{\frac{1}{2}m{u}^2}$
$=\frac{9\times 10^9\times 2\times 79{\left(1.6\times 10^{-19}\right)}^2}{5.5\times 1.6\times 10^{-19}\times 10^6}$
$=\frac{9\times 2\times 79\times 1.6\times 10^9\times 10^{-9}}{5.5\times 10^6}$
$=4.3\times 10^{-14}m$