16

The green light from copper flame is sadly not a single wavelength. It is very good question though because not many students even think about it. In order to "see" what wavelengths are present in a light source, you either need a spectroscope (old school way) or a monochromator with a detector (modern way). Of course you can search about Bunsen-...


7

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance. Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$...


3

The 'centrifugal' term appears when the coordinates are changed from $(x,y,z)$ to spherical $(r, \theta, \varphi)$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems. To summarize: the equation is $H\varphi(\...


3

$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$. As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$ then $$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$ For a massless object like a photon: $$p = \frac Ec$$. For an object with mass, there is ...


3

You are correct in your ideas. The effect is important in the van-der-waals interaction and appears as an additional $1/r^7$ term compared to the usual $1/r^6$ term. (This is sometimes called a retardation effect). It is present only at larger distances (a few nm) and arises only in induced-dipole interactions (dispersion interaction) because of the time it ...


3

Hmm, so I'm not entirely sure what notation you're using ($D_4$ sounded like Schönflies to me), but I'm sure you can figure out the equivalent. In general I agree with Andrew's point that molecular symmetry is determined by atomic positions and not orbitals. But at least to me as of now, this seems to be a thought exercise more than anything (after all, no ...


3

If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators. However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the ...


2

The tab sections Input, Examples and Basis Function Overview of Gen in Gaussian's online manual present, in tandem, a summary guide to what the various parameters listed in the question are and what the way they are ordered implies. Under the Input section, there is helpful text next to Defining a shell., which also has a useful pseudo-code block listing the ...


1

The potential energy between the O atoms in the H bond has the form of an anharmonic potential an example of the shape is the Morse potential or a Lennard-Jones 6-12. The energy is zero at large O-O separation and moves to a minimum as this distance is reduced (the point at which a the H bond exists with the H atom between the two O atoms, typically 0.15-0....


1

Let operators $\hat{A}$ and $\hat{B}$ commute, $[\hat{A},\hat{B}]=0$, and consider the eigenvalue equation for $\hat{A}$: $$ \hat{A}|\psi\rangle=\lambda|\psi\rangle. $$ The first thing to prove is that $\hat{B}|\psi\rangle$ is also an eigenstate of $\hat{A}$ with the same eigenvalue. To do this: $$ \hat{A}(\hat{B}|\psi\rangle)=\hat{B}(\hat{A}|\psi\rangle)=\...


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