If the nucleus 2712Al has nuclear radius of about 3.6 fm,

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If the nucleus ²⁷₁₂Al has nuclear radius of about 3.6 fm, then ¹²⁵₃₂Te would have its radius approximately as

It has been known that a nucleus of mass number A has radius
R = R₀A^1/3,
where R₀ = 1.2 × 10^–15 m
and A = mass number

In case of	27	Al,
	13

let nuclear radius be R₁ and for

	125	Te, nuclear radius be R₂
	32

For	27	Al, R₁ = R₀(27)^1/3 = 3R₀
	13

For	125	Te, R₂ = R₀(27)^1/3 = 5R₀
	32

	R₂	=	5R₀	=	5	R₁ =	5	× 3.6 = 6 fm.
	R₁		3R₀		3		3