25

It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within. There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells ...


18

This is due to the mass-energy equivalence and a phenomenon called binding energy. Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it. The binding energy of carbon-12 is ...


17

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of ...


11

No, nickel-62 is the most stable on a binding energy per nucleon basis. Fe-58 is second and Fe-56 is third. See Fewell, M. P., "The Atomic Nuclide with the Highest Mean Binding Energy", Am. J. Phys., vol. 63, pages 653-658.


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 ...


9

The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here. Now, the two reactions given are: \begin{array}{cc} \require{cancel} \ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\ \ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\ \end{array} Now using the rate law,...


8

The Wikipedia article on Nuclear Fusion starts off: In nuclear physics, nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons). The difference in mass between the reactants and products is manifested as the release of large amounts of ...


8

For energy levels in nuclei (and their gamma emission) the first place to look is at the Evaluated Nuclear Structure Data Files. I usually use the Brookhaven site, but there may be a closer mirror to you. Enter 24 into the Nuclide or Mass box and hit Enter. Check the 'Adopted Levels, Gamma' box by $^{24}$Mg, and request the pdf version (back up at the top). ...


7

There are different strategies for synthesizing superheavy elements. Two of the main strategies are referred to as "hot fusion" and "cold fusion". The first two reactions are good examples of the "hot fusion" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking. ...


7

There is a shell structure within the nucleus also. So the neutron/proton ratio varies from 1:1 for the light elements to 1.5 for the heavy elements. Note: I ripped off the image from another website. Not sure where they got it from...


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

Parallel or side reactions of the first order: Concept $$\require{cancel}\\ \ce{A ->[k_1] B} \ \ t=0\\ \ce{A ->[k_2] C} \ \ t=t$$ $$-\frac{\mathrm d[A]}{\mathrm dt}=k_1[A] + k_2[A] $$ $$-\frac{\mathrm d[A]}{\mathrm dt} = k_\text{eff} [A] \land k_\text{eff}=k_1+k_2$$ Effective order=1 $$\left(t_{1/2}\right)_\text{eff}=\frac {\ln 2}{k_\text{eff}} $$ $$\...


5

It has been done, not with bond formation but rather with bond breaking, and is called the Szilard–Chalmers reaction (credit to Loong for pointing out the name). When a biphasic mixture of ethyl iodide (with stable $\ce{^{127}I}$) and water is irradiated with neutrons, all the radioactivity from $\ce{^{128}I}$ is found as iodide in the aqueous phase; there ...


5

Yes, this really happened, probably (it's not a peer-reviewed effort). The reactor is a Farnsworth fusor, a very low-density, low-rate inertial-confinement reactor. They're surprisingly easy to build and get working; many people have done so, and verified their function by measuring neutron output. An accelerating voltage of 10 or 20 kV produces particle ...


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 {...


5

The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of ...


4

The stability of heavy nuclei is a function of both atomic number and atomic mass. Therefore, an appropriate number both of protons and of neutrons in the target nucleus must be attained, or else vanishingly short lifetimes can be expected. (See the Wikipedia article on, e.g., the island of stability, and links therein.) The 'source' nuclei are chosen to ...


4

I can't think of any examples of a nuclear reaction leading to a chemical reaction with bond formation. But you got me curious and I started wondering if the reverse process, i.e. chemical bond formation leading to a nuclear reaction might be possible. I couldn't think of any traditional bond formation reactions that would fall into this category. The ...


4

Actually they do collide. But, the emitted alpha particle carries much more energy than the binding energy of the electron(s) in a helium atom or $\ce{He^+}$ ion. Therefore such a collision scatters the electrons (which carry away some of the original energy of the alpha particle) instead of forming an atom.


4

The key here is that a system will tend towards the lowest energy state; in other words, for a process to be spontaneous, the final state must have lower energy than the initial state. The mass of a neutron is slightly higher than the combined mass of the proton, electron and neutrino that result from beta decay, meaning that the decay will be energetically ...


4

Just like electrons nuclei can exist in a number of quantised states. We see evidence of this when, during radioactive decay, a gamma ray is produced. This is a nucleus in an excited state decaying to the ground state, and the excess energy being radiated away as a gamma ray photon. This is precisely analogous to an electron in an atom falling from a high ...


4

The image is mislabeled: that red dot is 204 Hg, not 206 Pb. Plus, the text is wrong: the "last" (heaviest) stable isotope is not lead-206, but lead-208! The elements on the same vertical line of 206-Pb (which is 3 dots on the right of the mislabeled red dot) are 207-Pb and 208-Pb. The last dot on the top right is 209-Bi. Bismuth was found in 2003 to be ...


3

Spray perfume in the air. Ask the people around you to raise their hands when they can smell it. The people closest to you will raise their hands first. People farther away will raise then hands after that. A perfect example of diffusion.


3

As you mention most colored gases (Cl2, Br2, NO2, I2) are toxic and being a gas difficult to handle. Therefore not suitable for such demonstration. Maybe, you could try two small colored smoke bombs in a transparent container although the colour is not really a gas (more like fine particulate) in this case. I would use liquids with the most common diffusion ...


3

The photodisintegration of $\ce{^{9}_{4}Be}$ might come close to what you are looking for. When high energy photons excite a beryllium-9 nucleus it may decay into a neutron and two helium-4 nuclei ($\alpha$ particles). Neglecting the neutron, the latter are roughly half of the original nucleus. $$\ce{^{9}_{4}Be + \gamma -> 2 ^{4}_{2}He} + n$$


3

There are two reasons why nuclear reactions don't consider (or often account for) charge. The first is that charge is usually irrelevant to the reactions. If we are looking at changes in nuclei we can for all practical purposes ignore the electrons that would be associated with the nuclei under normal conditions. They are just irrelevant to the nuclear ...


3

In any nuclear reactions the sum of the mass numbers $A_i$ (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges $Z_i$ (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state: $$ \begin{align} \ce{^{222}_{86}Rn &→ ^{218}_{84}...


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