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An alpha -particle of energy 5 MeV is scattered through 180^@ by gold nucleus. The distance of closest approach is of the order of: |
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Answer» `10^(-12)cm` `r_(0)=(1)/(4pi espi_(0)). (2Ze^(2))/(1/2 mu^(2))` `=(9 xx 10^(9) xx 2 xx 79 xx (1.6 xx 10^(-19))^(2))/(5 xx 1.6 xx 10^(-13))` `=(9 xx 2 xx 1.6 xx 1.6 xx 10^(-16) xx 79)/(5 xx 1.6)` `=455 xx 10^(-16)m` `=4.55 xx 10^(-12)cm` |
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