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The ratio of the radii of the .(79)Au^(1...

The ratio of the radii of the `._(79)Au^(197)` nucleus to that of the radius of its innermost Bohr orbit is nearly `16 x`. Find the value of `x`. (Given `R_(0)=1.2xx10^(-15)m`).

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The correct Answer is:
6

`R=R_(0)A^(1//3)~~6.98xx10^(-15)m`
`r=(0.529xx10^(-10))/(Z)~~6.69xx10^(-13)m`
`(r)/(R)~~96= 16 rArr x=6`
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