If radius of the $Al presubscript 13 presuperscript 27$ nucleus is taken to be R_Al, then the radius of $Te presubscript 53 presuperscript 125$ nucleus is nearly

$open parentheses 53 over 13 close parentheses to the power of 1 third end exponent straight R subscript Al$
$5 over 3 straight R subscript Al$
$3 over 5 straight R subscript Al$
$open parentheses 13 over 53 close parentheses to the power of 1 third end exponent straight R subscript Al$

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Solution

5 over 3 straight R subscript Al

Radius space of space the space nucleus space is space given space by straight R space equals straight R subscript straight o straight A to the power of 1 divided by 3 end exponent space rightwards double arrow space straight R proportional to space straight A to the power of 1 divided by 3 end exponent straight R subscript Al over straight R subscript Te space equals open parentheses straight A subscript Al over straight A subscript Te close parentheses to the power of 1 divided by 3 end exponent space equals space open parentheses 27 over 125 close parentheses to the power of 1 divided by 3 end exponent space equals 3 over 5 straight R subscript Te space equals space 5 over 3 straight R subscript Al

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