## Z^2 in Schrodinger Equation

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

Bryce Ramirez 1J
Posts: 120
Joined: Sat Aug 24, 2019 12:16 am

### Z^2 in Schrodinger Equation

I'm confused by the books definition of trying to explain why Z^2 is more important than Z. The textbook says, "A skill to develop is to think about why a property depends on various parameters in a particular way. In this case you might wonder why the energy depends on Z 2 rather than on Z itself. The reason can be traced to two cooperating factors. First, a nucleus of atomic number Z and charge Ze gives rise to a field that is Z times stronger than that of a single proton. Second, the electron is drawn in by the higher charge and is Z times closer to the nucleus than it is in hydrogen. The two factors work together to give an overall energy lowering that is proportional to Z2." I don't understand the Z and Ze part of the first reason, and why the electron is Z times closer to the nucleus than if it were in a hydrogen.

Daniel Yu 1E
Posts: 100
Joined: Sat Aug 24, 2019 12:15 am

### Re: Z^2 in Schrodinger Equation

Z is the measure of the number of protons right? The radius of a helium atom is smaller than that of a hydrogen atom because there is a greater positive charge pulling the electrons tighter. Maybe that's what the book is trying to get at?

Siya Shah 1J
Posts: 50
Joined: Sat Aug 17, 2019 12:15 am

### Re: Z^2 in Schrodinger Equation

I believe what it is trying to say is that the strength of the nucleus pulling in the electrons is Z^2, which means that as the atomic number number increases, the nuclear charge Ze increases by the factor of Z^2.