If the nucleus ₁₃²⁷ Al has a nuclear radius of about 3.6 fm , then ₃₂¹²⁵ Te would have its radius approximately as:

Step-by-step solution

Nuclear radii (R= (R₀ ) A^ 1 / 3 ) where, (A ) is the mass number. ( R_ Te R_ Al = ( A_ Te A_ Al )^ 1 / 3 = ( 125 27 )^ 1 / 3 = ( 5 3 ) ) or, ( R_ Te = 5 3 R_ Al = 5 3 3.6=6 fm ). (Given (R_ Al =3.6 fm ))