Question

Mass number of a nucleus is \(\alpha\) and its radius is \(R_\alpha\). Radius of another nucleus of mass number \(\beta\) is \(R_\beta\). If \(\beta = 8\alpha\), then find \(\dfrac{R_\alpha}{R_\beta}\).

• A. \(\dfrac{1}{4}\)
• B. \(\dfrac{1}{2}\)
• C. \(\dfrac{1}{8}\)
• D. \(2\)

Hint:

Nuclear radius varies as the {cube root of mass number}: doubling the radius requires an {eightfold increase} in mass number.

Answer 1

The Correct Option is B

To solve the given problem, we first need to understand the relationship between the mass number of a nucleus and its radius. The formula that relates the radius of a nucleus \(R\) to its mass number \(A\) is given by:

R = R_0 A^{1/3}

where \(R_0\) is a constant.

Given that the mass number of one nucleus is \(\alpha\) and its radius is \(R_\alpha\), we have:

R_\alpha = R_0 \alpha^{1/3}

The other nucleus has a mass number \(\beta\) and its radius is \(R_\beta\). It is given that \(\beta = 8\alpha\). Substituting this into the radius formula, we get:

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

Simplifying the expression for \(R_\beta\), we have:

R_\beta = R_0 \cdot 8^{1/3} \cdot \alpha^{1/3}

We know that \(8^{1/3} = 2\), so:

R_\beta = 2 \cdot R_0 \alpha^{1/3}

Now, we need to find the ratio \(\dfrac{R_\alpha}{R_\beta}\):

\dfrac{R_\alpha}{R_\beta} = \dfrac{R_0 \alpha^{1/3}}{2 \cdot R_0 \alpha^{1/3}}

The \(R_0 \alpha^{1/3}\) terms cancel out, leaving:

\dfrac{R_\alpha}{R_\beta} = \dfrac{1}{2}

Thus, the correct option is \dfrac{1}{2}.