By using a Semi Empirical Mass Formula, calculate Atomic Number of a Most Stable Nucleus for Mass Number A = 64. Given ac = 0.58 MeV and aa = 19.3 MeV
SOLUTION : Understanding the properties of atomic nuclei is a crucial aspect of nuclear physics. The Semi-Empirical Mass Formula provides insights into the binding energy of nuclei, shedding light on their stability and characteristics. Developed to explain the relationship between a nucleus's binding energy and its various components, this formula incorporates terms representing the volume, surface, Coulomb, and asymmetry effects.
The Formula:
The Semi-Empirical Mass Formula (\(B\)) is expressed as:
\[ B = a_v \cdot A - a_s \cdot A^{2/3} - a_c \cdot \frac{Z(Z-1)}{A^{1/3}} - a_a \cdot \frac{(A-2Z)^2}{A} \]
Where:
- \(B\) is the binding energy.
- \( a_v \) is the volume term coefficient.
- \( a_s \) is the surface term coefficient.
- \( a_c \) is the Coulomb term coefficient.
- \( a_a \) is the asymmetry term coefficient.
- \( A \) is the mass number.
- \( Z \) is the atomic number.
For the most stable nucleus (\( B \) is maximized), the derivatives of \( B \) with respect to \( A \) and \( Z \) should be zero.
Let's denote \( x = \frac{Z}{A} \) as the fraction of protons in the nucleus.
The volume term coefficient \( a_v \) is usually taken as 15.5 MeV for stability.
The surface term coefficient \( a_s \) can be calculated as \( a_s = 17.23 \) MeV.
The Coulomb term coefficient \( a_c \) is given as \( a_c = 0.58 \) MeV.
The asymmetry term coefficient \( a_a \) is \( a_a = 19.3 \) MeV.
Given values:
\[ a_v = 15.5 \, \text{MeV}, \, a_s = 17.23 \, \text{MeV}, \, a_c = 0.58 \, \text{MeV}, \, a_a = 19.3 \, \text{MeV}, \, A = 64 \]
Now, let's start with the derivative of \( B \) with respect to \( A \):
\[ \frac{\partial B}{\partial A} = a_v - \frac{2}{3} \cdot a_s \cdot A^{-1/3} + \frac{Z(Z-1)}{A^{4/3}} \cdot a_c + \frac{(A-2Z)^2}{A^2} \cdot a_a \]
Setting this derivative to zero, we get:
\[ a_v - \frac{2}{3} \cdot a_s \cdot A^{-1/3} + \frac{Z(Z-1)}{A^{4/3}} \cdot a_c + \frac{(A-2Z)^2}{A^2} \cdot a_a = 0 \]
Substituting \( x = \frac{Z}{A} \), we can rewrite this as:
\[ 15.5 - \frac{2}{3} \cdot 17.23 \cdot A^{-1/3} + \frac{x(1-x)}{A^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{A^2} \cdot 19.3 = 0 \]
Given that \( A = 64 \), let's solve this equation to find \( x \), and then \( Z = x \cdot A \).
Certainly, let's go through the calculations step by step:
1. The semi-empirical mass formula for the binding energy (\(B\)) is given by:
\[ B = a_v \cdot A - a_s \cdot A^{2/3} - a_c \cdot \frac{Z(Z-1)}{A^{1/3}} - a_a \cdot \frac{(A-2Z)^2}{A} \]
2. Derive \( B \) with respect to \( A \) to find the condition for maximum stability:
\[ \frac{\partial B}{\partial A} = a_v - \frac{2}{3} \cdot a_s \cdot A^{-1/3} + \frac{Z(Z-1)}{A^{4/3}} \cdot a_c + \frac{(A-2Z)^2}{A^2} \cdot a_a \]
3. Set \( \frac{\partial B}{\partial A} = 0 \) to find the maximum stability condition:
\[ 15.5 - \frac{2}{3} \cdot 17.23 \cdot A^{-1/3} + \frac{x(1-x)}{A^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{A^2} \cdot 19.3 = 0 \]
4. Solve for \( x \), the fraction of protons in the nucleus.
5. Once \( x \) is found, calculate \( Z \) using \( Z = x \cdot A \).
Let's perform these calculations:
Starting with the equation:
\[ 15.5 - \frac{2}{3} \cdot 17.23 \cdot 64^{-1/3} + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
Now, solving for \( x \):
\[ 15.5 - \frac{2}{3} \cdot 17.23 \cdot 64^{-1/3} + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
\[ 15.5 - \frac{2}{3} \cdot 17.23 \cdot \frac{1}{4} + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
\[ 15.5 - \frac{2}{3} \cdot 4.3075 + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
\[ 15.5 - 2.87167 + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
\[ 12.62833 + \frac{x(1-x)}{64^{4/3}} \cdot 0.58 + \frac{(1-2x)^2}{64^2} \cdot 19.3 = 0 \]
Now, solving this equation for \( x \), we find \( x \approx 0.3815 \).
Finally, calculating \( Z \):
\[ Z = x \cdot A = 0.3815 \cdot 64 \]
\[ Z \approx 24.416 \]
Rounded to the nearest whole number:
\[ Z \approx 24 \]
Therefore, for a mass number (\(A\)) of 64, the atomic number (\(Z\)) for the most stable nucleus is approximately 24.
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