An alpha particle of kinetic energy 7.7 MeV is fired head-on at a gold nucleus (Z=79, A=197). Take e^2/(4piepsilon_0)=1.44 MeV fm and nuclear radius R=1.2A^1/3 fm. The distance of closest approach divided by the gold nuclear radius is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3as035

An alpha particle of kinetic energy $7.7 MeV$ is fired head-on at a gold nucleus ( $Z = 79$ , $A = 197$ ). Take $e^{2} / (4 π ϵ_{0}) = 1.44 MeV fm$ and nuclear radius $R = 1.2 A^{1/3} fm$ . The distance of closest approach divided by the gold nuclear radius is closest to

At closest approach K = (2)(79)(1.44 MeV fm)/d, so d = 227.52/7.7 = 29.5 fm. Gold radius R = 1.2 x

19 7^{1/3}

= 6.98 fm. Ratio = 29.5/6.98 = 4.2.