Question

In Rutherford experiment, a $ 5.3 \,MeV $ 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?
$ (\varepsilon_{0}=8.8\times10^{-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
$\frac{1}{2}mu^{2}=\frac{1}{4\pi\varepsilon_{0}} \frac{(ze) (ze)}{r_{0}}$
$\therefore $ Distance of closest approach
$r_{0}=\frac{1}{4\pi\varepsilon_{0}} \frac{2Ze^{2}}{\frac{1}{2}mu^{2}}$
$=\frac{9\times10^{9}\times2\times79\left(1.6\times10^{-19}\right)^{2}}{5.5\times1.6\times10^{-19}\times10^{6}}$
$=\frac{9\times2\times79\times1.6\times10^{9}\times10^{-9}}{5.5\times10^{6}}$
$=4.3\times10^{-14}m$