Question

Alpha particles of kinetic energy $ 7.7 \,MeV $ are being scattered by the nucleus of gold which has $ 79 $ electrons. The distance of closest approach of the alpha particles is
(Take $ \frac{1}{4\,\pi\, \varepsilon_{0} = 9 \times 10^{9} \, MKS} )$

• A. $ 4\times 10^{-14} m $
• B. $ 30 \times 10^{-15} m $
• C. $ 10\times 10^{-15} m $
• D. $ 7.9\times 10^{-14} m $

Answer 1

Answer: (B)

We know that
$\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{\left(Ze\right)\left(2e\right)}{r_{0}} = K$
$ 9\times 10^{9} \times\frac{ 79\times \left(1.6 \times 10^{-19}\right)^{2}}{r_{0}} \times 2 $
$ = 7.7 \times 10^{6} \times 1.6 \times 10^{-19}$
$r_{0} =\frac{9\times 10^{9} \times 79 \times\left(1.6\times 10^{-19}\right)^{2} \times 2}{7.7\times 10^{6} \times 1.6\times 10^{-19}} $
$ = 2.9\times 10^{-14} $
$r_{0} = 3.0\times 10^{-14} $
$ = 30\times 10^{-15} m$