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

Question #118814
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?
1
Expert's answer
2020-05-29T10:05:05-0400

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 \\\\\n 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 \\\\\nZ_{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} = \\\\\n= 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} = \\\\\n= 103.64 \\approx 104"

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


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