Question

A nucleus has mass number \(\alpha\) and radius \(R_\alpha\). Another nucleus has mass number \(\beta\) and radius \(R_\beta\). If \(\beta=8\alpha\), then \(R_\alpha/R_\beta\) is:

• A. \(1\)
• B. \(8\)
• C. \(0.5\)
• D. \(2\)

Hint:

Nuclear radius scales as the cube root of mass number: doubling radius requires eight times mass.

Answer 1

The Correct Option is C

To solve the problem given in the question, we need to use the nuclear physics formula that relates the radius of a nucleus to its mass number. This is expressed as:

\(R = R_0 \times A^{1/3}\)

where:

• \(R\) is the radius of the nucleus,
• \(R_0\) is a constant that is approximately equal to \(1.2 \, \text{fm}\) (femtometers), and
• \(A\) is the mass number (number of protons and neutrons).

We have two nuclei: one with mass number \(\alpha\) and the other with mass number \(\beta\). Given that \(\beta = 8\alpha\), we need to find the ratio \(R_\alpha/R_\beta\).

Substituting the mass numbers into the radius formula, we get:

• Radius of first nucleus: \(R_\alpha = R_0 \times \alpha^{1/3}\)
• Radius of second nucleus: \(R_\beta = R_0 \times (8\alpha)^{1/3}\)

We can simplify the second equation:

\(R_\beta = R_0 \times (8\alpha)^{1/3} \\ = R_0 \times 8^{1/3} \times \alpha^{1/3} \\ = R_0 \times 2 \times \alpha^{1/3}\)

Now, calculate the ratio:

\(\frac{R_\alpha}{R_\beta} = \frac{R_0 \times \alpha^{1/3}}{R_0 \times 2 \times \alpha^{1/3}} = \frac{1}{2}\)

Therefore, the ratio \(R_\alpha/R_\beta\) is \(0.5\).

So the correct answer is \(0.5\).