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An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional toA. `(1)/(ze)`B. `v^(2)`C. `(1)/(m)`D. `(1)/(v^(2))` |
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Answer» Correct Answer - C As we know, at the distance of closet approch is d. kinetic energy = Potential energy ` (1)/(2) mv^(2) =(1)/(4pi epsi_(0)) ((2e)(Ze))/(d)` where Ze = charge of target nucleus, 2e = charge of alpha nucles `(1)/(2)mv^(2)` =kinetic energy of alpha nuclesus of mass m moving with velocity v. or `d = (2Ze^(2))/(4pi epsi_(0)((1)/(2)mv^(2)))` `therefore " "d prop (1)/(m)` |
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