Myers mass formula to find the valley of stability for A=56

In summary, the formula for nuclear masses from MYERS AND SWIATECKI (1969) uses various coefficients and equations to calculate the energy of a nucleus. It is only valid for even-even nuclei and requires precise measurements of the neutron and proton masses. The conversation discusses the difficulty in obtaining the correct value for Z, the atomic number, using this formula. The similarity between the two calculated values for M(A,Z) suggests that the electron masses may also play a role.
  • #1
Leonardo Machado
57
2
Hi everyone.
I'm currently trying to master the use of the formula for nuclear masses from MYERS AND SWIATECKI (1969), https://www.sciencedirect.com/science/article/pii/0003491669902024.

$$
E=[-a_1+J\delta^2+0.5(K\epsilon^2-2L\epsilon \delta^2 +M\delta^4)]A+c_2 Z^2 A^{1/3}
+[a_2(1+2\epsilon)+Q \tau^2]A^{2/3}+a_3A^{1/3} \\
+c_1\frac{Z^2}{A^{1/3}} (1-\epsilon+0.5\tau A^{-1/3}) -2c_2 Z^2 A^{1/3} -c_3 \frac{Z^2}{A} -c_4 \frac{Z^{4/3}}{A^{1/3}},
$$

where

$$
I=(A-2Z)/A,\\
\delta=(I+\frac{3}{8} \frac{c1}{Q} \frac{Z^2}{A^{5/3}})/(1+\frac{9}{4}\frac{J}{Q} A^{-1/3}),\\
\epsilon=\frac{1}{K}(-2a_2A{-1/3}+L\delta^2+c_1\frac{Z^2}{A^{4/3}}),\\
\tau=(\frac{3}{2}\frac{J}{Q}\delta-\frac{1}{4}\frac{c_1}{Q}\frac{Z^2}{A^{4/3}}).
$$

and the coefficients are ( in MeV):

$$
a_1=15.677, \\
a_2=18.56, \\
a_3=9.34, \\
J=28.062, \\
K=294.8, \\
L=123.53, \\
M=2.673, \\
Q=16,04, \\
c_1=0.717, \\
c_2=0.0001479, \\
c_3=0.84, \\
c_4=0.5475.
$$

The point is, i can't obtain the correct Z that minimizes $$M(A,Z)=Z m_p + (A-Z) m_n + E(A,Z)$$. I'm obtaining Z=24 because M(56,24)=55.948365... and M(56,26)=55.948670..

Does anyone use this formula and know if there is any detail that I'm not seen?

*This formula only works to even-even nuclei.
 
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  • #2
What do you expect to get?
How many digits did you use for the neutron and proton mass?
The two numbers are extremely similar (so similar that the electron masses will play a role).
 

Related to Myers mass formula to find the valley of stability for A=56

1. What is the Myers mass formula?

The Myers mass formula is a mathematical equation used to predict the stability of an atomic nucleus based on its number of protons and neutrons. It was developed by physicist William Myers in the 1960s.

2. How does the Myers mass formula work?

The formula takes into account the binding energy of protons and neutrons in a nucleus, as well as the electrostatic repulsion between protons. It calculates the difference between the actual mass of a nucleus and the mass that would be expected based on the masses of its individual particles. If this difference is positive, the nucleus is predicted to be stable.

3. What is the valley of stability?

The valley of stability is a region on a graph of atomic mass versus atomic number where stable nuclei are located. Nuclei that fall within this region are predicted to be stable, while those outside of it are unstable and may undergo radioactive decay.

4. How does the Myers mass formula help find the valley of stability for A=56?

The Myers mass formula can be used to calculate the expected mass of a nucleus with A=56 (56 protons and neutrons). By comparing this expected mass to the actual mass of known nuclei with A=56, scientists can determine if these nuclei are stable or not. If they fall within the valley of stability, they are predicted to be stable.

5. What are the limitations of the Myers mass formula?

While the Myers mass formula is a useful tool for predicting the stability of nuclei, it does have some limitations. It does not take into account the effects of nuclear spin, which can affect the stability of some nuclei. It also does not account for the possibility of nuclear isomers, which have the same number of protons and neutrons but different energy states. Additionally, the formula is based on experimental data and may not accurately predict the stability of very heavy or very light nuclei.

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