1.

alpha- particle having 27 MeV has 1.14xx10^(-14) m distance of closest approach from a nucleus of atom. Calculate atomic no. of atom.

Answer»

100
103
105
90

Solution :PE, at `r_(0)` distance = kinetic energy,
`(K(Ze)(2e))/(r_(0))=27MeV`
`(kxx2Ze^(2))/(r_(0))=27MeV`
`:.Z=(27MeVxxr_(0))/(kxx2e^2)`
`=(27xx10^(6)xx1.1xx10^(-14))/(9xx10^(9)xx2xx1.6xx10^(-19))`
`=1.03125xx10^(2)=103`


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