If radius of the 2712Al nucleus is taken to be RAl, then

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If radius of the ²⁷₁₂Al nucleus is taken to be R_Al, then the radius of ¹²⁵₅₃Te nucleus is nearly:

As we know, R = R₀ (A)^1/3
where A = mass number
R_AI = R₀ (27)^1/3 = 3R₀
R_Te = R₀ (125)^1/3 = 5R₀

=	5	R_AI
	3

	5	R_Al
	3

	3	R_Al
	5

=		13		^1/3	R_Al
		53

=		53		^1/3	R_Al
		13