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The radius of n^(th) orbit of a single e...

The radius of `n^(th)` orbit of a single electron species is `0.132n^2A^@`. Identify the element.

Text Solution

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`r_(n)=(Kn^2)/(Z)`
`therefore 0.132n^2=(0.529xxn^2)/(Z)`
`therefore Z=(0.529)/(0.132)=4`
Since the atomic number is 4, the element is beryllium.
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