A 12.5 MeV alpha particle approaching a gold nucleus is deflected by 180 degrees. What is the closest distance to which it approaches the nucleus? Share with your friends Share 3 Pintu B. answered this Dear Student , The closest distance of approach is given by , d=14πε02e×ZeK.E=9×109×2e×79e12.5×1.6×10-19=9×109×2×79××1.6×10-1912.5=1.82×10-8 m Regards 0 View Full Answer Aayushi Katakwal answered this No links please :) -3
