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What is the distance of closest appr...

What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus ?

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using ` r_0 =(1)/( 4 pi epsi_ 0) (Ze^2 )/(E ) `, we get
`r_0 = ((9 xx 10^9 ) xx 79 xx (1.6 xx 10^(-19) )^2)/( 5xx 10^6 xx 1.6 xx 10^(-19) )=228 xx 10^(-11) m`
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