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If radius of the Al 1327 nucleus is taken to be RAl, then the radius of Te 53125 nucleus is nearly

a
35RAl
b
13531/3RAl
c
53131/3RAl
d
53RAl

detailed solution

Correct option is D

Radius of the nucleus R=R0A1/3∴  RAlRTe=AAlATe1/3 Here, AAl=27,ATe=125,RTe=?RAlRTe=271251/3=35⇒RTe=53RAl

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