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Consider an alpha-particle just in contact with .""_(92)^(238)U nucleus. Calculate the coulombic repulsion energy (i.e., the height of the coulombic barrier between ""^(238)U and the alpha-particle) assuming that the distance between them is equal to the sum of their radii. (r_(0)= 1.4 xx 10^(-13), e= 4.8 xx 10^(-10) esu) |
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Answer» SOLUTION :Distance between `U^(238)` and `He^(4)` nuclei, d, = radius of `U^(238)`+ radius of `He^(4)` `=1.4 xx 10^(-13) xx (238)^((1)/(3)) + 1.4 xx 10^(-13) xx (4)^((1)/(3))` `=1.0899 xx 10^(-12)cm` `therefore` COULOMBIC barrier `=(Z_(1)Z_(2) e^(2))/(d)` `=(92 xx 2 xx (4.8 xx 10^(-10))^(2))/(1.0899 xx 10^(-12))` `=3.899 xx 10^(-5)` erg |
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