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3

Due to the quantized energy levels of allowed electron orbitals, single atoms can easily absorb energy around certain narrow wavelengths. A cesium atom has one of these narrow absorption bands at a frequency of 9,192,631,770 Hz. A cesium clock can produce EM radiation in this region (microwaves) and detect how well the cesium atoms are absorbing it. So ...


3

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

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.


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 ...


0

In the semi-classical approach you treat the electromagnetic field of the incoming light as a (sinusoidal, with angular frequency $\omega$) perturbation $H^{\prime}$ in the Hamiltonian of the atom. The wavefunction of the atom can be expressed in terms of a linear combination of its eigenstates, each with a multiplying coefficient. If we start on the ground ...


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: ...


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 ...


0

Actually fission processes are stochastic, so we can't predict exact products of it. The most probable fission products are Cs-137 and Sr-90. The total amount of neutrons emitted is also quite unpredictable, but as I recall it usually lies in the range between 2 and 4.


-1

May be when number of proton and number of neutron exceeds the magic number like 2,8,20,50,82,126 etc by one ,the binding energy of extra nucleon is less than the average value. Which is known as first excitation energy.


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

Firstly, you cannot simply 'neglect' $a_0$ (as the *.pdf actually shows). In some test books the following substitution is used: $\rho=\frac{r}{a_0}$, so the $a_0$ doesn't have to be 'carried around'. As your expression for $P(\rho)$ is made of three factors, each factor can be evaluated for extrema individually. $\rho=0$ is an obvious minimum from the ...


2

The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two ...


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 ...


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 ...


0

Taking your questions in this order: As it is said energy tangled with mass and vice versa. Energy tangled with mass is the equation E = mc$^2$, meaning, as I am sure you know, energy and mass can convert into each other using this equation. For your particular question though, I would ask you to (temporarily) forget this idea as, although you are ...


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, ...


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

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

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 ...


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 ...


0

Well your reasoning is completely justified and valid. I am going to provide you with some elements to answer more what you want to know, but the exact response would depend on the case under study, and you will see why. Internal conversion happens mainly in heavy nuclei because they have electrons deeply bound, and their ionization energies are higher, ...


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 ...


0

Note: I'm aware this answer needs some work: Whilst I'm sure there will be a plethora of much more detailed answers I'll give a simple answer to question one. This is more about internally understanding how the process works than an extremely accurate description. Gamma ray is an electromagnetic wave. Basically we have a set of forces called electromagnetic ...


5

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 ...


-2

Why don't electrons collide among themselves Because they aren't anything like billiard balls. Check out the wave nature of matter. And take a look at the Wikipedia atomic orbitals article: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". The Heisenberg principle states that we ...


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 "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 ...


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. ...


9

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


0

If you crush hydrogen to very high densities, the Pauli exclusion principle will prevent electrons in the material from occupying the same quantum states. The electrons will obey Fermi-Dirac statistics and they will fully occupy states up to the Fermi energy - which increases as the density gets higher. Presumably in metallic hydrogen what happens is that ...


0

A short answer could be the following. Touching means experiencing some kind of reaction force, which is the macroscopic translation of the repulsive forces between the pen atoms and the object atoms. So, yes, you can touch object. On the other hand no damage is done to the atoms, as the coulomb forces that hold together atoms are much stronger than the ...


0

Let's analyze this problem. Very idealistically, the energy with which the pen of mass $m$ hits the object is given by $$ E = \frac{1}{2} I \omega^2,$$ where $I$ is the moment of inertia defined about an axis and $\omega$ is the angular velocity of the pen. Or even $$E = mgh,$$ where $g$ is the contribution from gravity and $h$ is the height from which it ...



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