aliz_khanz
Dec10-09, 01:23 PM
Question :- calculate the distance of closest apprach for a head on collision between 5.30Mev alpha particle and nucleus of a copper atom?
Answer
The distance of closest
approach is given by
d= _______qQ_______
4*pie*e0*K base alpha
= ____________2*29*(1.60*10^-19C)^2_______
4*3.142*8.85*10^-12C^2/Nm^2*5.30MeV*1.60*10^-13J/MeV
= 1.4 * 10^ -6 aNSWER..........
NOW I AM CONFUSED HERE ABOUT ONE THING AND ITS EATING ME ALIVE.....FROM WHERE DID 1.60*10^-13J/MeV CAME ? I HAVE USED COLUMB'S LAW !
Answer
The distance of closest
approach is given by
d= _______qQ_______
4*pie*e0*K base alpha
= ____________2*29*(1.60*10^-19C)^2_______
4*3.142*8.85*10^-12C^2/Nm^2*5.30MeV*1.60*10^-13J/MeV
= 1.4 * 10^ -6 aNSWER..........
NOW I AM CONFUSED HERE ABOUT ONE THING AND ITS EATING ME ALIVE.....FROM WHERE DID 1.60*10^-13J/MeV CAME ? I HAVE USED COLUMB'S LAW !