Answer to Question #118814 in Atomic and Nuclear Physics for maryam

Question

The Bethe-Weizsacker formula (Semi empirical mass formula) provides an excellent representation of the mass systematics of nuclei. Show explicitly that, for fixed A, M(A, Z) has a minimum value. Is there evidence for the "valley of beta stability"? What is the stablest nucleus with A = 16? What about A = 208?

Answer 1

1.The Bethe-Weizsacker formula:

$$m = Z m_p + (A - Z) m_n - \frac{1}{c^2} \left [ a_V A - a_{S} A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A - 2Z)^2} A \right ]$$

where $m_p, m_n, a_V, a_S, a_C, a_A$ are positive constants (can be taken from the https://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Calculating_the_coefficients).

Let's differenciate this expression wrt to Z and equate the derivative to zero:

$$\frac{dm}{dZ} = m_p - m_n - \frac{1}{c^2} \left [ - a_C \frac{2Z-1}{A^{1/3}} + a_A \frac{4(A - 2Z)} A \right ] = 0$$

Expressing Z from this equation:

$$- a_C \frac{2Z}{A^{1/3}} - a_A \frac{8Z} A = c^2( m_p - m_n) - a_C \frac{1}{A^{1/3}} - 4a_A \\ Z\left ( \frac{2a_C}{A^{1/3}} + \frac{8a_A} A\right ) = c^2( m_n - m_p) + \frac{a_C}{A^{1/3}} + 4a_A \\ Z_{min} = \frac{4Aa_A + Ac^2(m_n - m_p) + A^{2/3}a_C}{2a_C + 8a_A}$$

Let's check wheather this value gives us minimal m. The second derivative:

$$\frac{d^2m}{dZ^2} = \left ( \frac{2a_C}{A^{1/3}} + \frac{8a_A} A\right ) >0$$

Thus the $m(Z_{min}, A)$ is the minimal value indeed.

2. As far as $A = N+Z$, the obtained vlaue for $Z_{min}$ gives some region on the $(Z,N)$ graph with the mnimal value of m, which is called 'valley of beta stability'.

3. For A = 16 the stablest nucleus will be:

$$Z_{min} = \frac{4\cdot 16 \cdot 23.2 + 16\cdot c^2(938.57 MeV/c^2 - 938.3 MeV/c^2) + 16^{2/3}\cdot 0.71}{2\cdot 0.71 + 8\cdot 23.2} = \\ = 7.98 \approx8$$

Thus, the most stable nucleus at A = 16 is the nucleus of 16O

For A = 208:

$$Z_{min} = \frac{4\cdot 208 \cdot 23.2 + 208\cdot c^2(938.57 MeV/c^2 - 938.3 MeV/c^2) + 208^{2/3}\cdot 0.71}{2\cdot 0.71 + 8\cdot 23.2} = \\ = 103.64 \approx 104$$

Thus, the most stable nucleus at A = 208 is the nucleus of 208Rf.