**Since the parameter ****Δ****D _{i}**

**in the table**

**№ A-1**

**is given in terms of energy its formula looks like:**

** ****Δ****D _{i} = M_{i} ^{. }C^{2}– A_{i} ^{. }C^{2} **

**(3.12)**

**From the formula (3.12) we define value of weight in terms of energy ****M _{i} ^{. }C^{2}:**

** M _{i} ^{. }C^{2}= **

**Δ**

**D**

_{i}+ A_{i}^{. }C^{2}**(3.13)**

**Hence: ****Δ****D _{a} = M_{a} ^{. }C^{2}– A_{a} ^{. }C^{2}; **

** ****Δ****D _{b} = M_{b} ^{. }C^{2}– A_{b} ^{. }C^{2 }; **

** ****Δ****D _{c} = M_{c} ^{. }C^{2}– A_{c} ^{. }C^{2 };**

^{ }**M _{a} ^{. }C^{2}= **

**Δ**

**D**

_{a}+ A_{a}^{. }C^{2 };** M _{b} ^{. }C^{2}= **

**Δ**

**D**

_{b}+ A_{b}^{. }C^{2 };** M _{c} ^{. }C^{2}= **

**Δ**

**D**

_{c}+ A_{c}^{. }C^{2}**(3.14)**

**Let’s substitute formulas**** (3.14) in (3.3) and (3.4) or in (3.8): **

** E _{2с} = **

**ΔМ**

_{2с}^{. }**C**

^{2}=**M**

_{a}^{. }**C**

^{2}+**M**

_{b}^{. }**C**

^{2}–**M**

_{c}^{. }**C**

^{2}** ****E _{2с} = (**

**Δ**

**D**

_{a}+**A**

_{a}^{. }**C**

^{2}) + (**Δ**

**D**

_{b}+ A_{b}^{. }C^{2}) – (**Δ**

**D**

_{c}+ A_{c}^{. }C^{2})**As a result of condition (3.11), after reduction we shall receive**

** E _{2}_{с}**

**=**

**Δ**

**D**

_{a}+**Δ**

**D**

_{b}–**Δ**

**D**

_{c},**(3.15)**

**That corresponds with the formula**** (3.10).**

**After similar transformations with parameter E _{sv} we shall receive the formula**

** -E _{2}_{с}**

**= E**

^{sv}_{a}+ E^{sv}_{b}– E^{sv}_{c}**(3.16)**

**That corresponds to the formula**** (3.9).**

**For comparison, in the table**** №T-3.3, in lines 15, 15a, 15b; 16, 16a, 16b ****the results of calculations of allocated energy ****E _{2}_{с}**

**are resulted at**

**the moment synthesis of kernels of carbon**

_{17}

^{6}**C**

**and nitrogen**

_{17}

^{7}**N**

**,**

**by three ways**

**E**

_{2}_{с}**=**

**Δ**

**D**

_{a}+**Δ**

**D**

_{b}–**Δ**

**D**

_{c}

**– in lines 15,16;**

** ****E _{2с} = **

**ΔМ**

_{2с}^{. }**C**

^{2}=**M**

_{a}^{. }**C**

^{2}+**M**

_{b}^{. }**C**

^{2}– M_{c}^{. }C^{2}**– in lines 15a, 16a;**

**And**** -E _{2}_{с}**

**= E**

^{sv}_{a}+ E^{sv}_{b}– E^{sv}_{c}– in lines 15b, 16b.**From a comparison of the received data in lines 15, 15a, 15b; 16, 16a, 16b it is visible, that divergences in ****the results are significant, and in further calculations we shall use formulas**** (3.10)**** and (3.15). **

** (3.10)**

_{ }**At**_{ }_{ }**j=2**** ****E _{2}_{с}**

**=**

**Δ**

**D**

_{a}+**Δ**

**D**

_{b}–**Δ**

**D**

_{c}**(3.15)**

**It is possible to define the energy allocated ****in the time of synthesis of kernels or ****nuclides ****under ****the formulas:**

** (3.17) **

_{ }**In table**** № A-1 ****the data ****Δ****D is presented considering the weight of electronic environments of atoms, but at calculation of formulas**** (3.10)**** and (3.15) ****the weights of all ****electrons ****are completely reduced.**

** – Assumptions**

** We have defined formulas on which we shall count ****the allocated energy in time of synthesis of kernels. But for the continuation of the analysis it is necessary for us to make a number of assumptions since today it’s impossible to establish precisely which chains present in the synthesis in stars, how many kernels are participated in one act of synthesis, and under what conditions the quantity of kernels participating in one act varies.**

** Considering ****the wide scatter of stars weight, and the huge quantity of kernels participating in the synthesis in each star, it is possible to make a number of confident assumptions: **

** – Any variants, of chains of synthesis of kernels, without restriction are possible;**

** – Any quantity of kernels or ****nuclides ****participating in one act of synthesis from two j=2 and up to j=****A _{c} (A_{i})**

**is possible;**

** – It is possible that the synthesis of kernels with the mass numbers which are located at the end of the periodic table of elements and above A _{i }> 250, but unfortunately, we are limited by the knowledge in the limits of this table, therefore chains of the synthesis of kernels will be considered only by it.**

** For simplification of the maintenance of the analysis we accept, that ****during the time of synthesis, the sum of protons and neutrons of kernels ****a**** and ****b**** participated in the synthesis is equal to the number of protons and the neutrons of the received kernel ****c**** (a condition**** (3.11)).**

_{ }** – A choice of a way of the analysis**

**On an example of the synthesis of two kernels or ****nuclides ****we shall consider the chosen way of the analysis.**

** In the beginning it is necessary for us to choose a chain of synthesis.**

** Let’s take for an example a chain from hydrogen with ****Z****=1 and ****A****=1 up to helium with ****Z****=2 and ****A****=5, as shown in the scheme**** № S-3.1.**

**Chain of synthesis of a kernel of helium with ****Z****=2 and A=5 from two kernels or nuclides **

**Participating in one act of synthesis.**

**(13) Scheme № S-3.1. **

**Let’s make the table of ****nuclides ****and kernels participated in the synthesis of a kernel of helium ****He-5****.**

** (14)**** Table № T-3.1.**

**Parameters of nuclides and the kernels participating in synthesis of a kernel of helium He-5.**

**Let’s make formulas of synthesis under the scheme**** № S-3.1****:**

_{ }**1. _{ 1}^{1}H+_{1}^{0}n=_{2}^{1}H+E_{2H2}**

**Where ****E _{2H2}**

**– the energy allocated**

**in the time of synthesis**

_{1}

^{1}**H**

**and**

_{1}

^{0}**n**

**.**

** E _{2H2} = **

**Δ**

**D**

_{H1}+**Δ**

**Dn-**

**Δ**

**D**

_{H2}=7288.969+8071.5-13135.72=2224.749 KeV.** 2. _{2}^{1}H+_{1}^{0}n=_{3}^{1}H+E_{2H3}**

**Where ****E _{2H3}**

**– the energy allocated**

**in the time of synthesis**

_{2}

^{1}**H**

**and**

_{1}

^{0}**n**

**.**

** E _{2H3} = **

**Δ**

**D**

_{H2}+**Δ**

**Dn-**

**Δ**

**D**

_{H3}=13135.72+8071.5-14949.794=6257.426 KeV.** 3. _{3}^{1}H+_{2}^{1}H=_{5}^{2}He +E_{2He5}**

**Where ****E _{2He5}**

**– the energy allocated in the time of synthesis**

_{3}

^{1}**H**

**and**

_{2}

^{1}**H**

**.**

** E _{2He5}= **

**Δ**

**D**

_{H3}+**Δ**

**D**

_{ H2 }–**Δ**

**D**

_{He5}=14949.794+13135.72-11386.234=16699.28 KeV*.***Except for the value of energy allocated in one act of synthesis we interested in the value of the allocated energy in a step. The accessory of the reaction to one step or another is defined by the quantity of reactions of the synthesis which has passed the given kernel from a state when there were free protons and the neutrons included in its structure. **

**From the scheme**** № S-3.1**** it is visible, that the synthesis of kernel ****He-5 ****goes three steps.**

**At the first step two reactions of synthesis occurs**

** _{ 1}^{1}H+_{1}^{0}n=_{2}^{1}H+2224.749 KeV**

** _{ 1}^{1}H+_{1}^{0}n=_{2}^{1}H+2224.749 KeV**