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

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.