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If the nucleus of 13/27 has a nuclear radius of about 3.6 fm, then 52/125 would have its radius approximately as:

a. 1.8 fm
b. 2.4 fm
c. 3.6 fm
d. 4.8 fm

1 Answer

5 votes

Final answer:

Using the empirical formula for nuclear radius scaling with the cube root of its mass number, the nuclear radius of the nucleus with mass number 125 is approximately 4.8 fm.

Step-by-step explanation:

To find the nuclear radius of an atom with a different mass number A, we can employ the empirical formula r = ro A¹/³, where ro is the radius constant, approximately 1.2 fm, and A is the mass number of the nucleus. The radius of a nucleus scales with the cube root of its mass number. Given that the nucleus of 13/27Al has a nuclear radius of about 3.6 fm, we can calculate the radius of the nucleus of 52/125Te by using the formula.

First, we write the formula for 13/27Al:

  • rAl = ro (27)¹/³
  • 3.6 fm = ro (27)¹/³

Next, we use the same formula for 52/125Te:

  • rTe = ro (125)¹/³
  • rTe = (3.6 fm / (27)¹/³) × (125)¹/³
  • rTe ≈ 4.8 fm

Therefore, the correct answer is d. 4.8 fm.

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