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What is the distance of closest approach to the nucleus of an `alpha`partile which undergoes scattering by `180^(@)` is geiger-marsden experiment? |
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Answer» Correct Answer - `r_(0)=4.13`fm For closest approch, `(1)/(2)mv^(2)=K(Zexxe)/(r_(0))` For Rutherford experiment ltBrgt `(1)/(2)mv^(2)=5.5MeV=5.5xx10^(6)xx1.6xx10^(-19)J=8.8xx10^(-13)J` `8.8xx10(-13)=(9xx10^(9)xx2xx79xx(1.6xx10^(-19))^(2))/(r_(0))` `r_(0)=4.136xx10^(-15)m` `r_(0)=4.13` fm |
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