Hot answers tagged nuclear-physics
12
The answer is the there is some reduction in mass whenever energy is released, whether in nuclear fission or burning of coal or whatever.
However, the amount of mass lost is very small, even compared to the masses of the constituent particles. A good overview is given in the Wikipedia article on mass excess. Basically, the mass of a nucleus will in general ...
8
If you throw a bunch a uranium ore in one blob, nothing happens.
If you chemically purify the ore so that the only element present is uranium, still nothing happens.
The runaway chain reaction needed for a uranium-powered bomb involves U-235, an isotope having three fewer neutrons than the most common natural isotope U-238. According to Wikipedia,
The ...
5
From what I have read in "American Prometheus: The Triumph and Tragedy of J. Robert Oppenheimer" Teller was the first one to express this concern before the Trinity test.
Also quoting from: http://www.sciencemusings.com/2005/10/what-didnt-happen.html
Physicist Edward Teller considered another possibility. The huge temperature of a fission explosion -- ...
4
Yes, it will be harmful, because the isotope will travel through your bloodstream and deliver radiation damage to cells all over your body. If you want to know how harmful it is, check the recent case of Alexander Litvinenko, who was assassinated with polonium-210.
4
Short answer: A nuclear power plant contains a lot more nuclear material than an atomic bomb. The "Little Boy" bomb was detonated at 1968 feet (600m) over Hiroshima with the nuclear material dispersed quickly in the air; the Chernobyl meltdown contaminated its environment for decades.
Long answer:
http://en.wikipedia.org/wiki/Background_radiation
Total ...
4
While D-He3 fusion reaction rate peaks at smaller energies than D-D (see this picture), and produce more energy (18MeV for D-He3 vs. 3-4MeV for D-D reaction), this is not the main reason why some people think He3 is a 'better' fuel. The main reason is that D-He3 fusion fuel cycle is aneutronic. That is, all fusion products (if we disregard auxiliary ...
4
Well, if you search the internet it seems there are kids out there that make the claim of having built a fusion reactor . I watched this link. Note that in .56 minute he gives a small description, and does not claim breaking even, but that he demonstrated fusion. It is a plasma that he obviously creates and manages to fusion some deuterium that is not ...
4
Fusion happens between light nuclei. It cannot happen in room temperatures and pressures, it needs very high energies in order to strip the electrons from the nucleus and to overcome the electromagnetic repulsion of the positive charges .
The fusion reaction rate increases rapidly with temperature until it maximizes and then gradually drops off. The DT ...
3
The Weizsäcker formula and all similar formulae in nuclear physics are formulae for the masses of the nuclei, not atoms. That's true by design: the models behind the individual terms (droplet, shells etc.) are models for the nucleus only.
At the same moment, the accuracy of similar semiempirical formulae is not that marvelous which means that the errors are ...
3
Following through to previous news articles such as this one:
“This is my Inertial Electrostatic Confinement Fusion Reactor. It works on the property of inertial electrostatic confinement,” Conrad says.
See Wikipedia on inertial electrostatic confinement.
The actual design is a Farnsworth-Hirsch Fusor. Note Conrad's last name is Farnsworth which is a ...
3
Well, look at the other terms...in particular consider the Coulomb term as that one depends only on the proton number.
Walecka's book writes it as
$$ E_3 = \frac{3}{5} \frac{Z (Z-1)}{4\pi R_C} e^2 \approx a_3 \frac{Z^2}{A^{1/3}} \,.$$
This term is strictly positive and grows rapidly as the atomic number increases, while being slightly decreased by growing ...
2
Has your instructor (or your book) mentioned how much bigger a atom is than a atomic nucleus? On order of 10000 times.
Moreover, except for the $s$-shell electrons, most electron never come very close to the center (the $p$, $d$, etc shells all have nodes at $r=0$) so at the moment of fission the nuclei are sitting roughly at the middle of a roughly ...
2
It looks like Bi-209 is almost stable, with an extremely long half life, and its toxicity is low (http://en.wikipedia.org/wiki/Bismuth ). In general, however, alpha-radioactive materials are quite deadly when ingested (http://en.wikipedia.org/wiki/Polonium#Acute_effects )
2
Your basic nuclear reaction conserves the number of nucleons present.1
That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states.
So the only place available to get or lose energy in a reaction is by
Changing the flavor of nucleons. Every neutron converted to a proton gets you a ...
1
Is it possible to build a pure fusion and powerful nuclear bomb
"Powerful" is a relative term. According to the pure fusion weapon Wiki article (helpfully linked by Jim in the comments), a PFB with current technology would weigh about as much as the equivalent TNT yield. I suppose that would not be considered relatively powerful.
As is, I think, well ...
1
Well, the $Y_{lm}$ in the derivation above is actually a $Y_{lm}(\eta)$, so it is a function of $\eta$ and as such can be expanded in powers of $\eta$.
The author might only give a handwaving argument, but you can very rigorously prove that operators satisfying bosonic commutation relations, e.g., $[a^\dagger,a] = 1$, satisfy
$$[a^\dagger, f(a)] = ...
1
The first iteration -- fission driven fusion -- is exactly what happens in thermonuclear weapons. There the energy released by fission of heavy elements is used to fuse lighter elements (in practice isotopes of hydrogen, maybe with the help of lithium). In staged nuclear weapons design this fusion energy could be used to additionally drive the thermonuclear ...
1
The nuclear binding energy goes as displayed there. You can see that if you start with a heavy element (right hand side of the curve) and breaking it into 2 smaller elements (center of the curve), you end up with more energy. Say you want to break it up even more. Then the left part of the curve drops. It means that you will loose energy doing so. As a ...
1
Gugg's article is far enough as it covers everything you require. Firstly, It depends on the bomb. In other words, it's based on how bad the bomb is.
This Wiki article quotes it...
The amount of energy released by fission bombs can range from the equivalent of just under a ton of TNT, to upwards of 500,000 tons of TNT.
If that bomb is similar to ...
1
Nuclear physics often doesn't make sense without understanding the underlying quantum mechanics (and then, it often makes even less sense even though the numbers add up ;-)
In a nutshell, Deutrium is pretty stable compared to Helium-3. Helium-3 is more "desperate" to fill the missing neutron than Deutrium is to merge with a twin to form Helium.
In other ...
1
Your equations (2), (3) and (4) are all correct. I don't like the way (1) is written, because it relies on an unstated understanding that the $E$'s represent only mass energy, but with that understanding it is also correct.
This is just the conservation of energy in it's relativistic form where mass is just another kind of energy (previously you had to ...
1
The mass of the original polonium atom is 209.9828737(13)AU, while the mass of the lead atom is 205.9744653(13)AU and the mass of the helium is 4.00260325415(6)AU. The mass deficit gives you the amount of energy released.
1
Like dmckee says, the potential energy of electrons in an atom doesn't really compare to the energy of the nucleus. Since the nucleus is so tightly packed, and (in the case of Uranium) contains so many protons, they have a lot of potential energy—it takes a lot of work to "push" them together. The strong force holds protons and neutrons together when they're ...
Only top voted, non community-wiki answers of a minimum length are eligible

