# Tag Info

7

By adiabatic, I presume you mean the Born-Oppenheimer approximation (which is usually used). If you need a plot of potential against H-H distance, then you need to do a relaxed potential energy surface (PES) scan. ! UHF OPT def2-QZVPP %geom Scan B 0 1 = 0.3, 1.3, 30 end end The Scan directive in %geom section will perform a relaxed geometry scan (although ...

6

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$. There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, ...

4

Assume for a minute that instead of having a quantum oscillator with total energy $E$ you had a classical one with the same amount of energy. That energy is the sum of potential and kinetic contributions: $$E = E_\textrm{p}+E_\textrm{k}.\tag{1}$$ $E_\textrm{k}$ is zero and the total energy equal to the potential energy when the extension (amplitude) is at a ...

4

Option 2 is the most scientific and rigorous. You should re-run the IRC in solvent again because the reaction path can be different from what you got in the gas phase. Now, will it be different in solvent? Unless you are dealing with some exotic system, different reaction path with implicit solvation is unlikely (provided you have the same TS and minima with ...

4

Main differenece betweet STO's and GTO's is a $r$-expotetnt. Normalizing factor of GTO one can get from common procedure: $$N^2 \int_0^\infty \left(r^{n-1}e^{-\zeta r^2}\right)^2 r^2 dr =1$$ Of course, computation is not so simple for $n > 1$, but we can use online WolframAlpha (clickable calculation) and we get the formulas ...

3

I found an old slide from when I was teaching. These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($4\pi r^{2}R(r)dr$ if you understand the calculus). You can see that there's a blob of density that comes ...

2

You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.' To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised $$\int_0^\infty R_{n\ell}^*(r)R_{n\... 2 The hydrogen atom Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} +...

1

I suspect the problem is not mathematical. It is in the meaning of the wave function. Well. I will now show you how I explain it qualitatively in my high school classes. I state that the electron is like a violin string, but a string with three dimensions that is vibrating in the fourth dimension. As I know this last words are not understandable, I start ...

1

A broader point regarding functions that are finite: A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number. ...

1

In "real life", the hydrogen atom has its only electron in the s orbital. And unless it gets to an excitated state, that's the only thing that matters (whether you're studying the energy of the atom or the bounds it can make in terms of energy or symmetry). Nevertheless, the H atom is likely to have a spherical symmetry, so any of its p-orbitals ...

1

For the wave function of $\mathrm{s}$ orbital $\psi(r)$, the radial probability is: $P(r)=4\pi r^2 \cdot|\psi(r)|^2$. So even if the differential probability density $|\psi(r)|^2$ is nonzero for $r=0$, the integral probability over a spherical surface $P(r)$ in $r,r+\mathrm{d}r$ interval raises with $r^2$ and is zero for $r=0$. i understand the ...

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