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If the nuclear radius of ²⁷₁₃Al is denoted by R_(Al), then the radius of ¹²⁵₅₃Te is approximately:
- (5)/(3)R_(Al)
- (3)/(5)R_(Al)
- ((13)/(53))^(1/3)R_(Al)
- ((53)/(13))^(1/3)R_(Al)
Correct answer: (5)/(3)R_(Al)
Solution
Nuclear radius R = R0 * A^(1/3), so R_Te/R_Al = (125/27)^(1/3) = 5/3. Hence R_Te = (5/3) R_Al.
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