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Whatis thedistanceof closestapproachwhena 5MeVprotonapproachesa goldnucleus ? |
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Answer» Solution :USING` r_0=(1)/( 4 pi epsi_ 0) (Ze^2 )/(E ) `,we get `r_0 = ((9 xx 10^9 ) xx 79xx (1.6 xx 10^(-19) )^2)/( 5XX 10^6 xx 1.6 xx 10^(-19) )=228 xx 10^(-11) m` |
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