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Why does nuclear size vary linearly with atomic mass number?
Nucleus contains nucleons viz; protons and neutrons. Nucleus is not just pile of balls (neutrons and protons) as is generally depicted in many cases. And we don't know much about nucleus. Obviously linear variation of size with atomic mass number is experimented fact. What are theoretical explanations for this?

The number of atoms in $1kg$ of carbon is given by $=\frac{12\times <!--\; \times \; -->1000}{N_A}$, but if I remember correctly the number of atoms is equal to the number of moles times the Avogadro's number and hence should be $=\frac{1000\times <!--\; \times \; -->N_A}{12}?$? It does a similar thing for the number of atoms in $1kg$ of Uranium $-<!--\; -\; -->235$, which is given to be in the solution $=\frac{235\times <!--\; \times \; -->1000}{N_A}$, and not $=\frac{1000\times <!--\; \times \; -->N_A}{235}$
So which is the correct answer?

Isotope, One of the two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behaviour but with different atomic masses and physical properties. every chemical element has one or more Isotopes. 3 examples of isotopes. for example what these mass numbers denote before further progress?

$B=(Z\times mH+N\times mn-<!--\; -\; -->mA)c^2$ , where $mH$ is the mass of Hydrogen, mn is the mass of the neutron and $mA$ the mass of the nucleus with atomic number $Z$, mass number $A$ and $N$ neutrons.
What is the mass defect for real?

Why is the relationship between atomic number and density not linear? What are the factors that affect the density of an atom?

"Each such orbital can be occupied by a maximum of two electrons, each with its own spin quantum numbers."
I though that all electrons had same spin quantum number $s=1/2$, being the difference the $z$-component of the angular momentum, $m_s\in <!--\; \in \; -->\{-<!--\; -\; -->1/2,1/2\}$.
I'm confusing the nomenclature ?

I have to calculate the Bethe-Bloch equation for a $\mathrm{C}{\mathrm{O}}_2$ absorber, I know this is a stupid question, but is Atomic number additive? Or do I need to do something a little more complex?I was think $Z=22$ since $C$ has $6$ protons and $O$ has $8$, is that correct?

Is this equation for the mass defect is correct
$\mathrm{\Delta <!--\; \Delta \; -->}m=Z\ast <!--\; \ast \; -->m_p+Z\ast <!--\; \ast \; -->m_e+N\ast <!--\; \ast \; -->m_n-<!--\; -\; -->A$
where
- $Z$ - Atomic number
- $A$ - Mass number
- $N$ - number of neutrons
- $m_p$ - mass of the proton
- $m_n$ - mass of the neutron
- $m_e$ - mass of the electron
I came up with another formula also
$\Delta M=[(Z(m_p)+N(m_n))-(M)]$
where $m$ is the observed mass
Which one is correct and what is the difference between both these equations?

The photon here does not interact with the nucleus; it just needs the nucleus for momentum conservation. Why does the screening effect then effect pair production cross section? Also, why does the cross section increase with atomic number?

To state the title question perhaps more precisely:
What is the largest photon energy $E_{\mathrm{\gamma <!--\; \gamma \; -->}}$ and the corresponding mass number $A$ and atomic number $Z$ of a suitable nucleus ${}_Z^{A}X$ (presumably in a ground state) such that the hypothetical reaction
${}_Z^{A}X+\mathrm{\gamma <!--\; \gamma \; -->}\to <!--\; \to \; -->{}_{(Z-<!--\; -\; -->z)}^{(A-<!--\; -\; -->a)}Y+\text{whatever remains (with combined charge}+z\text{)}$
is "kinematically" forbidden for any values $1\le <!--\; \le \; -->a<A$ and $Z\ge <!--\; \ge \; -->z\ge <!--\; \ge \; -->Z+a-<!--\; -\; -->A$,while conforming to the standard model?

Trying to look for a derivation of the Rutherford scattering formula from Hyperphysics, but cannot find one. How is this equation derived?
$N(\mathrm{\theta <!--\; \theta \; -->})=\frac{N_{i}nL{Z}^2{k}^2{e}^4}{4r^{2}K{E}^2{\sin }^4<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}/2)}$
where
$N_i$ is the number of incident alpha particles
$n$ is the number of atoms per unit volume in the target
$L$ is the thickness of the target
$Z$ is the atomic number of the target
$e$ is the electron charge
$k$ is Coulomb's constant
$r$ is the target-to-detector distance
$KE$ is the kinetic energy of the alpha particle
$\mathrm{\theta <!--\; \theta \; -->}$ is the scattering angle

When calculating the orbital energy of an electron in a hydrogen-like atom, the orbital velocity is required. However, to derive this value, the equation given is:
$Z{K}_e{e}^2/r^2$ and $K_e$
is the atomic number. However, there are plenty of systems that behave with an electron orbital which are non-atomic, like those have an quantum potential well. In these cases, there is no single defined atomic number, so how would one go off calculating this?

While studying NMR theory, my textbook explained that only nuclei with odd mass numbers are NMR active because they have non-integer spin quantum numbers and nuclei with an even mass number and atomic number have I=0 spin.
This doesn't make much sense to me. NMR works by generating a magnetic field which interacts with spinning charges (like a proton or electron) that also generate a magnetic field. So how come the nucleus of Carbon-12 is "invisible" to NMR?
In other words, if we treat each proton as miniature bar magnet that is capable of flipping between alpha and beta states (parallel and antiparallel), what is stopping the protons from "reversing spin" in a carbon-12 nucleus?

Equation is the definition of specific heat given in terms of the fermi-dirac distribution function $f$ and the density of states equation $g$
$C_e(T_e)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\mathrm{\partial <!--\; \partial \; -->}f(\mathrm{\epsilon <!--\; \epsilon \; -->},\mathrm{\mu <!--\; \mu \; -->},T_e)}{\mathrm{\partial <!--\; \partial \; -->}{T}_e}g(\mathrm{\epsilon <!--\; \epsilon \; -->})\mathrm{\epsilon <!--\; \epsilon \; -->}d\mathrm{\epsilon <!--\; \epsilon \; -->},$
The first consideration that I make is that the fermi level is a function of temperature, so I use the following conservation equation to solve for the fermi level
$N_e={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}f(\mathrm{\epsilon <!--\; \epsilon \; -->},\mathrm{\mu <!--\; \mu \; -->}(T_e),T_e)g(\mathrm{\epsilon <!--\; \epsilon \; -->})d\mathrm{\epsilon <!--\; \epsilon \; -->}.$
where $N_e$ is the number of free electrons. For example in Copper it would be 1 electron/atom.
My question is around how to determine the number of free electrons in a metal as a function of temperature. All the analysis that I have seen just assumes a static amount of free electrons for a material, but we know as $T\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ the number of free electrons goes towards the atomic number of the metal.
My question is how can I solve for $N_e(T)$ if I know $N_e(T_0)$?

Given that the principal quantum number, $n$, has a value of $k\in <!--\; \in \; -->\mathbb{N}$, is there a closed form equation for calculating the number of atomic orbitals in terms of $k$?
As an example, for $n=2$, the number of atomic orbitals is $4$. But can this value be calculated for any $n=k$?

How do nuclear properties influence the atomic properties and vice versa?

Finding new stable superheavy elements is big interests in nuclear physics.
Nuclides with $Z>92$ are not found in nature, but can be made artificially.
Usually these nuclides become more unstable as their atomic number increases.
But, the island of stability predicts there are some stable isotopes of superheavy elements.
My question is why finding stable superheavy element is important?
Is it just for academic curiosity? Or is there some important application?

What would be the atomic number of the atom (may be hypothetical) whose 1s electron moves at $0.99c$ (the speed of light)?
Quantum mechanics might have an answer, but I do not know the necessary maths to calculate.

How does adding one more particle to the nucleus of an atom give that atom new properties? I can see how it changes it's mass, that's obvious... But how does it give that new atom different properties like color?
A good example would be: start with a copper atom (Cu), with the atomic number 29, thus Cu has 29 protons, and you add one proton to the nucleus you are left with an atom of Zinc (Zn) with the atomic number 30, thus 30 protons. The first element mentioned is a totally different color than the second, and conducts electricity better etc.
Not only protons, but neutrons, which are the same type of particle (Baryon) affect the properties of the element in a much different and much less important manner. Adding a neutron only creates an isotope of that element, not a different one all together, unlike adding a proton.

As the atomic number $Z$ increases, is the Hartree-Fock approximation getting better and better, or worse and worse?