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12

A few years ago the XUV physics group at the AMOLF Institute in Amsterdam were (to my knowledge the first to be) able to directly image the orbitals of excited hydrogen atoms using photoionization microscopy. For more details see the paper, Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States. A.S. Stolodna et al. ...


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The first images of hydrogen s orbitals were obtained in 2013 by physicists in the Netherlands.


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There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...


4

When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for $J=\frac{1}{2}$ with $L=0$, and those are $J_z=\pm\frac{1}{2}$ (Or some linear combination of them) As to why this is, ...


4

The peaks at $E = 1099\:\mathrm{keV}$ ($P = 56.5\:\%$) and $E = 1292\:\mathrm{keV}$ ($P = 43.2\:\%$) are the most important gamma lines for Fe-59. If a significant Fe-59 activity is present in your sample and your detector is sufficiently sensitive in this energy range, you should be able to see both peaks. The intensity of both peaks might be reduced due to ...


3

The answer is already on page 2 of your link above: "Among the large number of radionuclides of medical interest, Sc-44 is promising for PET imaging. Either the ground-state Sc-44g or the metastable-state Sc-44m can be used for such applications, depending on the moleculeused as vector." So the metastable state Sc-44m decays to the ground state Sc-44g.


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Be-7 is common atmospheric radionuclide produced by cosmic ray spallation of nitrogen and oxygen. Ground level concentration of Be-7 is in order of ~mBq per cubic meter of air. Main deposition process of Be-7 is a wet scavenging which yields to ~Bq per litre of rainwater. It is therefore possible to find Be-7 in background (depends on location of ...


3

Fission reaction produce a distribution of several different nuclei. Wikipedia (User:JWB) has a nice graph that shows the relative probabilities. However, in the case of a single reaction, where it is given that a Cs-137 nucleus is produced, you can probably be more specific because there are correlations between the fission products, but it will require ...


2

As indicated in the answer to this physics.stackexchange question the total amount of nucleons is preserved during fission. As a result the atomic number of a daughter product can be predicted if another was already known. In the case of $U_{92}^{235}$ fissioning to $Cs_{55}^{137}$, we know the atomic number of the second fission product is given by: ...


2

If the slit is much smaller than the wavelength, there will not be significant path-length difference between beams passing through different points inside the slit. If the slit is much larger than the wavelength, most of the beam passing through the slit will not be affected by the edges of the slit. I also guess that Born had in mind Fraunhofer diffraction ...


2

When a nucleus decays, there is usually excess energy (binding energy of the nucleons) liberated, and this energy is carried away by any of the "fragments" - daughter nuclei, emitted particles (alpha, beta, gamma, neutrino, ...). The amount of energy involved can be quite large (think atom bomb), so when the energy liberated is released in the form of ...


2

Because atoms do not show the linear Stark effect. As atoms have no permament electric dipole moment the leading order effect is the quadratic Stark effect, which is supressed by another factor of the field strength (which is already small compared to the atomic field strengths). Early precision spectrometry was based entirely on atoms in electric discharge ...


2

The "bullet" in Thin Man was a conical shell, which would go *over" the target, reducing contact time incredibly. It was molded on an old cone clutch design from an automobile, with two conical rings of plutonium meeting at high speed. The problem was contamination. In effect, they would have to cast the core pieces, install, and drop within 28 days, else ...


2

A typical implosion assembly of a Pu pit is time limited by the maximum detonation velocity of the high explosive used. As a ball park figure, you are not going to get a shock a compression rate of more than 10 km/s because of that limit. So in theory a gun assembly might work if you could accelerate the projectile to twice that velocity (the implosion is ...


2

The occurrence of pair production within the detector depends on gamma energy and detector material. The occurrence of escape peaks in addition to a given full-energy peak depends on detector geometry and sample geometry. If the gamma energy is large enough to make pair production relevant, the photon may disappear and be replaced by an electron and a ...


1

The Fermi level in the Richardson-Dushman equation is already the Fermi level at the temperature your system has. You do not have to account for thermal excitations seperately. Usually the equation is applied with a high temperature and is intended for just this.


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According to wikipedia, Vanadium-48 decays via $\beta^+$ (positron emission) to Titanium-48, which is a stable isotope. The emission of neutrons for Vanadium-48 isn't allowed becuase it doesn't conserve energy: Vanadium-48 has a mass of 47.9522537 u, and Vanadium-45 plus 3 neutrons have a total mass of 44.965776 u +3ยท1.00866491600 u = 47.991770 u. For a ...


1

First of all let us make clear the the Bohr model posits orbits, with some imposed by hand quantization constraints, as the stationary assumption. Orbitals for the hydrogen atom are the solutions of the Schrodinger equation with the hydrogen potential energy and helped develop quantum mechanical theory. In QM the solution of the schrodinger equation is a ...


1

Typically the order of magnitude of angular spread of a fringe in a pattern can be given by lambda/Width. If lambda is much smaller than width the diffraction pattern will be difficult to see. In the other case of lambda much greater than the width the entire screen would be covered by just the first fringe.


1

Maybe you should look up the definition of what an orbital is again. Your question doesn't make much sense. https://en.wikipedia.org/wiki/Atomic_orbital#Shapes_of_orbitals The "Shapes of Orbitals" section should help the understanding. I suppose by orbital you mean an atomic orbital, which is basically nothing else than a designated area in which the ...



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