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If the nucleus AI 1327 has a nuclear radius of about 3.6 fm. then Te 52125 would have its radius approximately as

a
6.0 fm
b
9.6 fm
c
12.0 fm
d
4.8 fm

detailed solution

Correct option is A

If R is the radius of the nucleus, the corresponding volume 43πR3 has been found to be proportional to A.The relationship is expressed in inverse form as R=R0A1/3The value of R0 is 1.2 x 10-15 m, i.e., 1.2 fmTherefore RAIRTe=R0(AAI)1/3R0(ATe)1/3RAIRTe=(AAI)1/3(ATe)1/3=(27)1/3(125)1/3=35or RTe=35xRAI=35x3.6=6 fm.

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