# Tag Info

52

The white flashes seen in the video are the result of radioactive contamination of the original film. Airborne radioactive particles were deposited onto the film and caused a local overexposure of the negative, quite similar to the historic experiment of Antoine Henri Becquerel. That's why the flashes only appear in individual spots in individual frames of ...

26

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 ...

16

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.

10

Due to its very long half-life, U-238 is not particularly radioactive; the specific activity is only about 12 kBq/g. The emission probability of gamma radiation is less than 0.1 % per disintegration of U-238. Furthermore, the gamma energy is relatively low. Therefore, the direct gamma radiation from pure U-238 (i.e. disregarding Th-234 and further decay ...

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 ...

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...

7

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 ...

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 ...

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 {... 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 The concerned silver nuclide with 61 neutrons is \ce{^108_47Ag}. This nuclide with 47 protons and 61 neutrons lies in the so-called valley of β-stability. Image taken from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002), p. 42 Nuclides on right side of the valley (higher neutron numbers) are unstable to decay by β− ... 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 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 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 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$$3 In any nuclear reactions the sum of the mass numbers A_i (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges Z_i (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state:$$ \begin{align} \ce{^{222}_{86}Rn &→ ^{218}_{84}...

3

Note that binding energy of nucleons does not mean contained energy, but the energy needed to break it. Therefore kernels with lower binding energy per nucleon contain more nuclear energy per nucleon and have higher mass per nucleon. It means forming kernels with higher binding energy per nucleon releases energy and vice versa. It is the same as for ...

3

Let's look at just the carbon-14 atoms. Assune 18.5% of your body mass is carbon and you weigh 80 kg. One carbon atom in a trillion is carbon-14. Working out the resulting mass of carbon-14 atoms in grams, dividing by 12.01 g/mol and multiplying by Avogadro's Number leads to roughly $7.4×10^{14}$ atoms. Carbon-14 has a half-life of about 5730 years. ...

2

Nuclear decay is accelerated (artificially produced) all the time. Per wikipedia, U-238 is not usable directly as nuclear fuel, though it can produce energy via "fast" fission. In this process, a neutron that has a kinetic energy in excess of 1 MeV can cause the nucleus of 238U to split in two. So a neutron with 1 MeV kinetic energy can split the ...

2

No. When an atom decays, it splits into two smaller atoms, and its protons are divided between those two new atoms. The smallest number of protons you can split off an atom is one, and a single proton by itself is still a hydrogen atom. Note that you can split a proton into sub-atomic particles, but this would not be called "decay" anymore.

2

The hydrogen nuclei are spaced out by about $\pu{70pm}$, if I guessed/remembered the number correctly. To generate a helium nucleus, this distance has to be reduced to almost $\pu{0pm}$ against the repulsion of like charges. So if you just bombarded a hydrogen molecule with neutrons from one side, nothing would happen since the two would not fuse. It may ...

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