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49

Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the ...


14

Nuclear reactions appear to violate both the Laws of Conservation of Mass and Energy because mass is converted into energy or vice versa. However, the concept of mass-energy equivalence that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the Law of Conservation ...


12

For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved. $$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$ $$Z_1 = Z_2 + Z_3$$ $$A_1 = A_2 + A_3$$ Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\...


10

The given expression $\ce{^{16}_{8}O}\,(\mathrm{n},\alpha)\,\ce{^{13}_{6}C}$ translates as $\ce{^{16}_{8}O ->[+\ \mathrm{n}][-\ \alpha] ^{13}_{6}C}$ or $\ce{^{16}O + n -> ^{13}C + ^4He}$ and the expression $\ce{^{9}_{4}Be}\,(\alpha,\mathrm{n})\,\ce{^{12}_{6}C}$ translates as $\ce{^{9}_{4}Be ->[+\ \alpha][-\ \mathrm{n}] ^{12}_{6}C}$ ...


9

Light can pass through a gold foil though, it just has to be thin enough. Pure gold is a very malleable substance and can be beaten with a hammer into foils of around 100 nm thickness. Sources suggest the gold foil used in the Geiger–Marsden experiment (known more commonly as the Rutherford gold foil experiment) was about 86 nm thick. Somewhere around this ...


9

There are various designs and materials for control rods, which are more technically correctly called control assemblies. The choice depends on the type and purpose of the reactor, the size of the core, the availability of other means for power control (e.g. boric acid in the reactor coolant, burnable poison rods, $\ce{UO2-Gd2O3}$ fuels, recirculation pumps, ...


9

There are fewer decays because there are fewer atoms to decay The simple reason why the number of decays (strictly, the number of decays per unit time) decreases in simple radioactive decay is because there are fewer atoms left to decay. Nuclear decay is probabilistic. The probability of any given unstable atom decaying is constant (independent of time or ...


8

The radiocarbon method is in principle fairly simple. $^{14}\ce{C}$ is an isotope of Carbon that has a relative abundance of around $10^{-10} \%$ on earth. Addition by @MaxW: $^{14}\ce{C}$ is formed in the upper atmosphere from $^{14}\ce{N}$ so that the relative amount of $^{14}\ce{C}$ in the atmospheric $\ce{CO_2}$ is reasonably constant. Since ...


7

By beta decay in the nucleus of the atom a neutron decays into a proton, an electron, and an electron antineutrino and a lot of energy. It will lose a little mass, since $E = mc^2$, so the decay energy is coming from the mass we lose. The electron (or beta particle) has a high energy and leaves the atom. $$\ce{_0^1n -> _1^1p+ + _{0}^0e- + \overline{\nu_{...


7

Some of the electrons will react with positrons formed during the fusion processes; they annihilate each other and give off high-energy radiation (also called gamma rays): Source But this process happens at the core, and there is no way that those electrons get out to the surface. So my best guess is that the electrons which don't get annihilated just ...


6

Researchers are looking for the Island of Stability, a predicted group of heavy nuclei with halflives in the minutes or even days. While we have some understanding that certain nuclear configurations and compositions are more stable, we are currently unable to reliably to a priori predict the stability, the half-life, or even the decay mechanisms of a ...


6

An "element" must be defined as the set of all atomic nuclei having a specified number of protons. Definitions based on electrons (or other leptons) can't be used because how many electrons are associated with an element changes with the atom's environment. Defining an "atomic nucleus" as a set of protons and neutrons, in a common nuclear potential well, ...


6

To clear up what are becoming confusing comments. An atom is a nucleus of protons and neutrons surrounded by electrons. An atom can decay by fission to make two or more atoms with a smaller number of protons and neutrons in each new atom. The electrons get distributed between them. Now in physics we'd normally describe this as a nuclear decay, rather ...


6

There is no theory, nor is there a need for one. To put in in other words: yes, there is a theory, and that's the one you are already familiar with. Antimatter elements are precisely the same as ordinary matter elements, only the other way around. There "are" (read: "could be", though some were actually detected) antihydrogen, antihelium (all with the same ...


5

We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to ...


5

Naively, the nuclear electric field at Z ~ 137 or greater, reciprocal of the Fine Structure constant, would "spark the vacuum." Vacuum would be torn into electron-positron pairs. The electrons go in to convert protons to neutrons plus neutrinos. As stated above, non-classical treatment suggests we'll never get near a cold nucleus that sparks the vacuum. ...


5

An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about nuclei, not the entire atom. The reaction is to be read as "A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino." Also, note that the neutrino has no charge.


5

Here's a list of types of radioactive decay. The most notable types of decay that are not among the classic three involve the direct emission of a free proton or a neutron, the emission of atomic clusters other than helium nuclei (alpha particles), the absorption of one of the innermost shell electrons into the nucleus, or spontaneous fission of the unstable ...


5

The nuclear shell model is a useful "first approach" to determining nuclear spin. It doesn't always work, but it is a relatively simple way to make a first attempt. Here is a nuclear shell energy diagram. As you can see it is somewhat analogous to an electron orbital energy diagram. A set of shells are filled by neutrons and a separate set is filled by ...


5

Nuclear fission involves the splitting of a heavy nucleus into two lighter fragments plus the release of energy. As an example here is one route by which $\ce{^235U}$, a fissile material, can split apart. $$\ce{^{235}_{92}U + ^{1}_{0}n -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + 200 MeV}$$ Note the 3 neutrons generated in the fission process, they ...


5

This is really a physics question but I will answer it anyway. I'm not sure how much subatomic physics you know so I will give two different versions. Simply put, a neutron decays to form a proton and electron. $$\ce{n -> p+ + e-}$$ This explains why the proton number increases by one to form Xenon. More properly a down quark inside a neutron decays to ...


5

Since the gravitational force between two protons is negligible there must be another force holding the nucleus together. This is the strong nuclear force, which as the name suggests is extremely strong but it is also extremely short range and so it's effects are only felt on the scale of nuclei and baryons. As you can see in the graph, if two protons ...


5

I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct? Not exactly. The atom as a whole emits or absorbs the photon. There is no reason to single out the electron versus the nucleus in such transitions. Can atomic nuclei eject photons? Yes there are two ways a nucleus in particular (as ...


5

Reflect means to bounce back, as in a ball bouncing off a wall. The average density of an atom is very low, so the observed reflection was startling... like throwing many billiard balls at a mass of fluffy cotton candy and every now and then a ball comes bouncing back! The explanation offered was that there must be some "hard", dense object inside ...


5

Consider that to get diffraction effects you generally need the wavelength of the diffracting particle to be related to a length scale of importance in your sample. For the moment, take it to be something on the order of an inter-atomic spacing, roughly an Angstrom ($\pu{\mathring A}$). Taking an alpha particle at $\pu{3MeV}$ (the ones originally used were ...


5

Neutrinos come from the elementary nuclear reactions: $$n^0\to p^++e^-+\overline\nu$$ or $$p^+\to n^0+e^++\nu$$ Not sure what do you mean by collision, since beta decay normally does not include any collision (unless there is an electron capture, of course; but neutrino emission occurs anyway, capture or no capture). As for the purpose, they have none, much ...


5

It is a general principle, not limited to nuclear chemistry, but is common for many areas, e.g. for the reaction kinetic of the 1st order. All processes, where the value time rate is proportional to the value, have value time evolution in the form of the exponential function. $$\frac {\mathrm{d}x}{\mathrm{d}t}= -k \cdot x$$ leads to $$x= x_0 \cdot \exp {...


4

Further to @ManishEarth's answer, note that the distinction is important in doing mass-energy calculations. In the above example, you can find the energy released by subtracting the tabulated mass of a $N^{14}$ atom from the tabulated mass of the $C^{14}$ atom, and ignore the electron mass. Why? Because the actual Nitrogen atom produced only has 6 orbital ...


4

Since this thread seems to have been bumped by @Community a better answer is in order. Physicists don't typically make a distinction between rest mass $m_0$ and moving mass $m$ but rather consider mass to be the invariant of the energy-momentum $4$-vector: $mc^2=\sqrt{E^2-\boldsymbol{p}^2c^2}$ where $m$ is the invariant mass of the system, $E$ its total ...


4

In the context of photochemistry, radiation time or half life period are not used, as far as I know. The only relevant technical term i'm aware of -in that context- is half-life (time), which describes the time needed for the concentration of an excited species to decrease to 50% of the initial value, cf. S. E. Braslavsky, Glossary of terms used in ...


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