Question

Let the binding energy per nucleon of a nucleus be denoted by \( E_{\text{b/n}} \), and the radius of the nucleus is denoted by \( r \). If the mass numbers of nuclei A and B are 64 and 125 respectively, then:

• A. \( r_A<r_B \)
• B. \( r_A>r_B \)
• C. \( E_{\text{b/n}}^A>E_{\text{b/n}}^B \)
• D. \( E_{\text{b/n}}^A<E_{\text{b/n}}^B \)

Hint:

The radius of a nucleus increases with the cube root of its mass number, and the binding energy per nucleon decreases with increasing mass number. Use these relationships to solve problems related to nuclear properties.

Answer 1

The Correct Option is A

We are given that the mass numbers of nuclei A and B are 64 and 125 respectively, and we need to determine the relationship between the binding energy per nucleon and the radius of the nucleus for these two nuclei. Step 1: Relation Between Radius and Mass Number The radius \( r \) of a nucleus is related to the mass number \( A \) by the empirical formula: \[ r \propto A^{1/3} \] where \( A \) is the mass number of the nucleus. This relationship tells us that the radius of a nucleus increases with the cube root of its mass number. For nuclei A and B, we have the mass numbers: - \( A_A = 64 \) - \( A_B = 125 \) Using the relation for the radius, we can express the ratio of the radii of the two nuclei as: \[ \frac{r_A}{r_B} = \left( \frac{A_A}{A_B} \right)^{1/3} = \left( \frac{64}{125} \right)^{1/3} \] Calculating the value: \[ \frac{64}{125} = 0.512 \] \[ \left( 0.512 \right)^{1/3} \approx 0.799 \] Thus: \[ r_A<r_B \] Step 2: Relation Between Binding Energy per Nucleon and Mass Number The binding energy per nucleon, \( E_{\text{b/n}} \), is related to the mass number \( A \) by the empirical formula: \[ E_{\text{b/n}} \propto \frac{1}{A} \] Thus, the binding energy per nucleon decreases with increasing mass number. For nuclei A and B, we can express the ratio of the binding energies per nucleon as: \[ \frac{E_{\text{b/n}}^A}{E_{\text{b/n}}^B} = \frac{A_B}{A_A} = \frac{125}{64} \approx 1.953 \] Thus, the binding energy per nucleon of nucleus A is greater than that of nucleus B, as \( E_{\text{b/n}}^A>E_{\text{b/n}}^B \). Step 3: Conclusion From the above analysis: - \( r_A<r_B \) - \( E_{\text{b/n}}^A>E_{\text{b/n}}^B \) Thus, the correct answer is: \[ \boxed{(A) \, r_A<r_B} \]