Question

If \(r_1\) and \(r_2\) are the radii of the atomic nuclei of mass number 4 and 32, respectively, then the ratio \(\left(\dfrac{r_1}{r_2}\right)\) is

• A. 1:2
• B. 1:3
• C. 1:4
• D. 1:5

Hint:

Use \(R = R_0 A^{1/3}\), so \(r_1/r_2 = (4/32)^{1/3}\).

Answer 1

The Correct Option is A

Step 1: Idea:
Nuclear matter has constant density. Constant density means the mass number \(A\) grows in the same way as volume, so radius grows as the cube root of \(A\).

Step 2: Set up the ratio:
Let the volume be $V = \frac{4}{3}\pi r^3$. With equal density, $V_1/V_2 = A_1/A_2 = 4/32 = 1/8$.
Also $V_1/V_2 = (r_1/r_2)^3$.

Step 3: Solve for the ratio:
So $(r_1/r_2)^3 = 1/8$.
Try $r_1/r_2 = 1/2$: its cube is $1/8$, which matches.
So $r_1/r_2 = 1/2$.

Step 4: Compare with the options:
A ratio of 1:2 is the first option. The ratios 1:3, 1:4 and 1:5 give volume ratios of 1/27, 1/64 and 1/125, which do not equal 1/8, so they are wrong.

Final Answer:
The nuclear radius ratio is 1:2. \[\boxed{r_1 : r_2 = 1 : 2}\]