## Bindings in helium #### Binding energy

Theory of atomic nucleuses has proved that binding energy between particular nucleons does not depend on their charge, i.e. that it approximately holds:
E(N,N)=E(N,P)=E(P,P), where N is neutron and P proton.

Often reported reason is, that nucleuses with inverse number of protons and neutrons (e.g. isotopes H(3,1),(p=1,n=2) and He(3,2),(p=2,n=1)) are similar and their bindings energy differs only by Coulomb repulsion of protons.

Now let us do one test.

#### Example

Let us assume, we can compute bindings energy with this formula:

E(p,n) = c(P,P)*E(P,P) + c(P,N)*E(P,N) + c(N,N)*E(N,N)
where p,n is number of protons P and neutrons N and c(P,P), c(P,N), c(N,N) number of bindings between them.

Bindings energy of the four simplest atomic nucleuses is:

```  X(A,Z)   p  n   E(p,n)  c(P,P) c(P,N) c(N,N)
---------------------------------------------
H (2,1)  1  1   2.224    0      1      0
H (3,1)  1  2   8.482    1      2      0
He(3,2)  2  1   7.718    0      2      1
He(4,2)  2  2  28.295    1      4      1
```

We get three equations:

```  E(1,1) = 0*E(P,P) + 1*E(P,N) + 0*E(N,N) = 2.224
E(1,2) = 0*E(P,P) + 2*E(P,N) + 1*E(N,N) = 8.482
E(2,1) = 1*E(P,P) + 2*E(P,N) + 0*E(N,N) = 7.718
```
Hence:
```	 E(P,P) = 3.270
E(N,N) = 4.034
E(P,N) = 2.224

Check:
E(1,1) =  1*2.224           = 2.224
E(1,2) =  2*2.224 + 1*4.034 = 8.482
E(2,1) =  2*2.224 + 1*3.270 = 7.718
```

Binding energy of E(2,2) is then:

```	 E(2,2) = 1*E(P,P) + 4*E(P,N) + 1*E(N,N)
E(2,2) = 1*3.270  + 4*2.224  + 1*4.034 = 16.200.
```
It is - in comparison with real energy of He(4,2) - very low
(the difference 28.295- 16.200= 12.095 is c. 3*E(N,N)).

#### Test

Let us complete the formula for E(p,n):

E(p,n) = c(P,P)*E(P,P)+c(P,N)*E(P,N)+c(N,N)*E(N,N)+p*E(P)+n*E(N) ,

where

```  E(1,1) = 0*E(P,P) + 1*E(P,N) + 0*E(N,N) +1*E(P)+ 1*E(N) = 2.224
E(1,2) = 0*E(P,P) + 2*E(P,N) + 1*E(N,N) +1*E(P)+ 2*E(N) = 8.482
E(2,1) = 1*E(P,P) + 2*E(P,N) + 0*E(N,N) +2*E(P)+ 1*E(N) = 7.718
E(2,2) = 1*E(P,P) + 4*E(P,N) + 1*E(N,N) +2*E(P)+ 2*E(N) =28.295
```
If E(N)=E(P) we get:
```	  E(P)   = -6.048
E(N)   = -6.048
E(P,P) = -2.778
E(N,N) = -2.014
E(P,N) = 14.319

Check:
E(1,1) = 14.319 -6.048 - 6.048 = 2.224
E(1,2) = 2*14.319 -2.014 -3* 6.048 = 8.482
E(2,1) = -2.778 + 2* 14.319 -3*6.048 = 7.718
E(2,2) = -2.778 + 4* 14.319 -2.014 -4*6.048  = 28.295
```

The binding energy E(P,P) is nearly the same as E(N,N), but not the same as energy E(P,N).
Nucleuses with inverse number of protons and neutrons confirm or disprove nothing - they have the same number of bindings E(P,N).