Atomic weight in respect to the binding energy?

Question

My book says that the weight of helium (with the nucleon number of 4 and proton of 2) is that of $6,6447*10^{-27}$ kg. Earlier the book stated that if the proton number is left out it means that the atom is neutral included the electrons. Does this imply that the weight of the electrons are included in tables if the proton number is excluded?
http://en.wikipedia.org/wiki/Isotopes_of_helium#Table
I can't find the helium isotype with the proton number written out on the list. If I take the one with nucleon number 4 the weight would be
$4.00260325415 \times 1,66 \times 10^{-27} = 6,644321402 \times 10^{-27} $ kg
Is this correct? If I do subtract for two electrons it doesn't add up either. It's just off by a tiny bit.

I need to find the weight correctly because I'm using this to estimate the binding energy. Because that is suppose to be equal to the difference.

Answer 1

The atomic number $Z$ is the same as the number of protons. Thus, if you are talking about Helium ($He$), the proton number is $2$.

The isotopic mass (not weight!) is the mass of an atom of the isotope, reflecting all of its constituents and the (negative) binding energy. In your example, you would write the mass of $^4_2He$ as:

$$ M(^4_2He) = 2m_p+2m_n+2m_e-U$$ where $U$ is the binding energy of the whole atomic system, including the nucleus.

As for your calculation, using $^4He$ isotopic mass: $$4.00260~u = 6.646478 \cdot 10^{-27}~kg$$ You don't have to subtract anything, this is the actual mass of the atom. If you want to know the value of the binding energy, you will need to add up the masses of the constituent protons, neutrons and electrons, and evaluate the difference.

This page may be helpful reading on the subject: http://education.jlab.org/qa/pen_number.html

Also, I can't find any version of Helium with atomic mass $6.6447\cdot10^{−27}~kg$ that you quote in the beginning, but note that it looks alot like the atomic mass of $^4He$ at $6.64647\cdot10^{-27}kg$, but with the middle $6$ left out. There is quite possibly a typo in your book which has led to this confusion.

As a check that you are doing the calculation properly, you should find a binding energy for $^4_2He$ of about $0.03~u$.