- #1
aliz_khanz
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Problem For Structure of Nuclei -- Enigmatic !
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 !
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 !