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If radius of the 27/13 Al nucleus is taken to be R_Al, then the radius of 125/53 Te nucleus is nearly:

1. 5/3 R_Al
2. 3/5 R_Al
3. (13/53)^(1/3) R_Al
4. (53/13)^(1/3) R_Al

Correct answer: (53/13)^(1/3) R_Al

Solution

The radius of a nucleus is proportional to the cube root of its mass number (A). Thus, the ratio of the radii of two nuclei is proportional to the cube root of the ratio of their mass numbers. For 125/53 Te and 27/13 Al, the ratio is (53/13)^(1/3).

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