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In a head on collision between an alpha particle and gold nucleus (Z=79), the distance of closest approach is 39.5 fermi. Calculate the energy of `alpha`-particle. |
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Answer» Here, Z=79, `r_0=39.5fermi=39.5xx10^(-15)m , E=?` form `E=1/2mv^2=1/(4pi in_0)((Ze)(2e))/(r_0)` `E=(9xx10^9xx79xx2(1.6xx10^(-19))^2)/(39.5xx10^(-15))` `=(18xx79xx2.56)/(39.5xx10^(-14))J` `=(9.216xx10^(-13))/(1.6xx10^(-13))=5.76MeV` |
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