Question

Alpha particles of kinetic energy $7.7MeV$ are being scattered by the nucleus of gold which has $79$ electrons. The distance of closest approach of the alpha particles is
(Take $\frac{1}{4\pi {\epsilon }_0=9\times 10^{9}MKS})$

• 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
$\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\left(Ze\right)\left(2e\right)}{r_0}=K$
$9\times 10^9\times \frac{79\times {\left(1.6\times 10^{-19}\right)}^2}{r_0}\times 2$
$=7.7\times 10^6\times 1.6\times 10^{-19}$
$r_0=\frac{9\times 10^9\times 79\times {\left(1.6\times 10^{-19}\right)}^2\times 2}{7.7\times 10^6\times 1.6\times 10^{-19}}$
$=2.9\times 10^{-14}$
$r_0=3.0\times 10^{-14}$
$=30\times 10^{-15}m$