1.

An `alpha`-particle having `K.E. =7.7MeV` is scattered by gold `(z=79)` nucleus through `180^(@)` Find distance of closet approach.

Answer» `K.E.=7.7MeV=7.7xx10^(6)xx1.6xx10^(-19)J=1.23xx10^(-12)J`
`(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)`
Using we get `: (9xx10^(9)xx2xx79xx(1.6xx10^(-19))^(2))/(1.23xx10^(-12))`
`r_(0)=3xx10^(-14)m`
From the above example it is clear that nuclear diamension cannot be greater than `3xx10^(-14)m`


Discussion

No Comment Found

Related InterviewSolutions