Question:medium

The radius of an atomic nucleus of mass number \(64\) is \(4.8\;fermi\). Then the mass number of another atomic nucleus of radius \(6\;fermi\) is

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Nuclear radius varies with mass number as \[ R=R_0A^{1/3} \] Therefore, \[ A\propto R^3 \]
Updated On: Jun 22, 2026
  • \(64\)
  • \(81\)
  • \(100\)
  • \(125\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Recall the nuclear radius law.
The radius of a nucleus depends on its mass number through \[ R = R_0 A^{1/3} \] so $R \propto A^{1/3}$.
Step 2: Set up the ratio for the two nuclei.
Comparing the unknown nucleus with the known one, \[ \frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} \]
Step 3: Insert the given values.
With $A_1 = 64$, $R_1 = 4.8\ \text{fermi}$ and $R_2 = 6\ \text{fermi}$, \[ \frac{6}{4.8} = \left(\frac{A_2}{64}\right)^{1/3} \]
Step 4: Simplify the left side.
Since $\dfrac{6}{4.8} = 1.25$, \[ 1.25 = \left(\frac{A_2}{64}\right)^{1/3} \]
Step 5: Cube both sides.
Raising to the third power, \[ (1.25)^3 = \frac{A_2}{64} \quad\Rightarrow\quad 1.953 = \frac{A_2}{64} \]
Step 6: Solve for the mass number.
Thus $A_2 = 1.953 \times 64 \approx 125$, giving \[ \boxed{125} \]
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