Problem For Structure of Nuclei - Enigmatic

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SUMMARY

The discussion centers on calculating the distance of closest approach for a head-on collision between a 5.30 MeV alpha particle and a copper nucleus. The formula used is derived from Coulomb's Law, specifically d = (qQ) / (4πε₀K), where q and Q represent the charges involved. The conversion factor of 1.60 x 10^-13 J/MeV is clarified as the fundamental charge expressed in joules per MeV, essential for converting energy units in this context. The final calculated distance of closest approach is 1.4 x 10^-6 m.

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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 !
 
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its the conversion between MeV and joules. Both are energy units.
 


aliz_khanz said:
FROM WHERE DID 1.60*10^-13J/MeV CAME ? I HAVE USED COLUMB'S LAW !

It's the fundamental charge, e, expressed in unusual units.
 

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