Question #118571
the semi-empirical formula provides an excellent representation of the mass systematic of nuclei. show explicitly for fixed A, M has the minimum value. is there any evidence for 'valley of beta stability' what is the stablest nucleus with A=16
1
Expert's answer
2020-05-28T13:32:43-0400

1.The Bethe-Weizsacker formula:


m=Zmp+(AZ)mn1c2[aVAaSA2/3aCZ(Z1)A1/3aA(A2Z)2A]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 mp,mn,aV,aS,aC,aAm_p, m_n, a_V, a_S, a_C, a_A (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:


dmdZ=mpmn1c2[aC2Z1A1/3+aA4(A2Z)A]=0\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:


aC2ZA1/3aA8ZA=c2(mpmn)aC1A1/34aAZ(2aCA1/3+8aAA)=c2(mnmp)+aCA1/3+4aAZmin=4AaA+Ac2(mnmp)+A2/3aC2aC+8aA- 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:


d2mdZ2=(2aCA1/3+8aAA)>0\frac{d^2m}{dZ^2} = \left ( \frac{2a_C}{A^{1/3}} + \frac{8a_A} A\right ) >0

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


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


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


Zmin=41623.2+16c2(938.57MeV/c2938.3MeV/c2)+162/30.7120.71+823.2==7.988Z_{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}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.


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