New answers tagged

0

The energy of a nucleon in a nucleus is not an observable quantity (think about how you would measure it), and so is entirely dependent on the theoretical model used to describe the nucleus. Once nucleons are bound into a nucleus, they don't retain their individual identity. There is just "the nucleus". In fact, one can --with a suitably large computer--...


6

Your existing answer talks about quark confinement, but stable nuclei can't really be described using quark and gluon degrees of freedom. Also your existing answer doesn't answer your title question: why don't nuclei collapse to a point? To first approximation, nuclei do collapse into a point. The diameter of a nucleus is typically about $10^{-5}$ the ...


0

http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/imgnuk/bcurv.gif The answer to your question is contained in the curve linked above. Details: As can be seen from the binding energy curve, the binding (per nucleon) increases sharply as more nucleons are added at the lower mass numbers; it reaches a maximum at the Iron group (mass number near 56); and ...


18

First, the strong force acts on scales where our classical idea of forces as something that obeys Newton's laws breaks down anyway. The proper description of the strong force is as a quantum field theory. On the level of quarks, this is a theory of gluons, but on scales of the nucleus, only a "residual strong force", the nuclear force remains, which can be ...


0

If you have a Poisson distribution with a probability $$P(n)=\frac{\lambda^n}{n!}e^{-\lambda}$$ that there will be $n$ events per bin, then $\lambda$ is the mean number of events per bin. You can get this via a direct calculation, $$ ⟨n⟩ =\sum_{n=0}^\infty nP(n) =\sum_{n=0}^\infty n \frac{\lambda^n}{n!}e^{-\lambda} =\lambda e^{-\lambda}\sum_{n=1}^\infty \...


1

Producing ultra-heavy elements in nature is not easy. So their absence "in nature" does not mean they cannot exist or cannot be created given the right conditions. Some details: The valley of stability becomes increasingly n-rich, so neutron capture reactions are essential. To get beyond lead requires rapid neutron capture in the r-process. The ...


2

Because no rational process can make them. I've been over the tables, and there are only a couple of possible reactions to get there for any nuclei, and they require two rare ones. Alpha particle capture just isn't going to cut it. Look at the curve; you need more neutrons. We remember that all elements heavier than iron have primary sources as neutron star ...


2

As you say, it is because of the nuclear force. The charges do repel each other, but not strongly enough to overcome this other, stronger force. However, this is why in nature the neutrons in a substance tend to outnumber the protons--the neutrons can provide additional nuclear force without adding to the electrical repulsion. Having only protons isn't ...


3

The atomic states win in a landslide. The reason is that the coulomb force is long ranged while the nuclear force is short ranged. That means that the number od discrete nuclear states is finite while the number of discrete atomic states is infinite. Even for a neutral atom where you might think that cancellation of charges would shorten the interaction ...


3

Griffith's 1987 book correctly states a totally reasonable hypothesis, that the neutron's core is positive. Here's a simple model which probably goes back to Fermi: the neutron ought to spend part of its time as a virtual proton-$\pi^-$ pair, in a strong-interaction analog to the photon spending part of its time as an electron-positron pair; since the ...


0

Ernest Moniz and John Kerry wrote an op-ed in the Washington Post about the recent nuclear deal with Iran. In it, they say that even though inspections can be delayed for 24 days, that delay is not a problem .My question is: how does this detection work, and how hard is it to defeat? Let us have the info as to how the monitoring of Nuclear establishments/...


1

The limit on the strong force holding things together is given by the combination of a range and the degeneracy of the nuclear matter. The residual strong force is well described mathematically by a Yukawa potential $$ V(r) = - \frac{g^2}{4 \pi c^2} \frac{e^{-mr}}{r} \,,$$ where the mass $m$ that appears in there is roughly the pion mass and $g$ is an ...


3

The reason why not any nuclei (for e.g. one which has only protons) can exist in nature is due to the fact that they will have a very low Binding Energy, compared to a nuclei which has the same number of nucleons, but with a more stable proton/neutron ratio. A simple model which allows us to get fairly accurate estimated of Binding Energies is the Liquid ...


2

The question uses the term "Usually" which is not a correct description , however the decay schemes can be understood by analzing the process in detail. An alpha particle is identical to a helium nucleus, being made up of two protons and two neutrons bound together. There are models in which a nucleus can be seen as cluster of alpha-particles; say Carbon -...


1

Recall that the value and sight of $\mathbf{r} \times \mathbf{p}$ angular momentum depends on the point around which you chose to measure it. A free particle can (indeed, does) have angular momentum around any and all points relative which it is moving and has a non-zero impact parameters. But, frankly, that's not a very interesting statement.


1

I worked on a program two decades ago where we were to determine if nations "XYZ" were building nuclear weapons based on intelligence "ABC." Before I did this I took a DOE course that amounted to intermediate level (210) nuclear weapons. A lot of data on this is classified, and one needs CWDI Q-clearance to know this stuff. A lot of effort has gone into ...


0

In this context (as in many other contexts) the mass of a nucleus is incorrectly but very commonly also called its "weight". This is not precise, but in many cases you can just let it slide... In this case, however, the answer to your question has not so much to do with the mass of the nucleus than with the nuclear binding energy of the nuclei. The nuclear ...


0

Lamarsh is a great textbook, but a little dated and out-of-print. You can usually pick up copies on E-Bay. The book currently used in many junior-level nuclear engineering courses is E. E. Lewis, "Fundamentals of Nuclear Engineering", Elsevier (2008). ISBN 978-0-12-370631-7


2

The minimum radius of a spherical object to be a black hole is given by : r = 2Gm/ (c^2) From this, I think we may be able to calculate the minimum density for the object to be a black hole, which is: d = (21/704)((c^6)/((G^3)(m^2)) (Assuming pi = 22/7) it is, d = (7.37 x 10^79 )/(m^...


0

I haven't run the numbers, I assume JR is correct. I do know that the 'density' of a proton is much less than that required (isn't this obvious?) for gravitational collapse to occur (otherwise, it would). The only thing I want to add here is a cautionary note about our lack of understanding about quantum gravity. That is, as soon as you want to discuss ...


41

Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high. The Schwarzschild radius of a black hole is given by: $$ r_s = \frac{2GM}{c^2} $$ and for a mass of $1.99 \...


0

Answering this question is one of the major successes of 20th-century physics. For strong decays, Gamow's alpha-tunneling model is quite successful. It relates the lifetime of an alpha emitter to the energy released in the decay using the approximately-valid assumption that nuclear density is constant and that the nucleus has a relatively sharp edge. For ...


-3

There is a thought experiment that might help the theoretician. Think of the nuclear attractive force as dynamic spatial field and the C14 atom as binding the two extra neutrons with this force field. Extremely rarely (avg. $4700$ years) there is weak spot in the containment field aligned with the neutron, and it escapes. You can build a simple 2D ...


6

W.u stands for Weisskopf unit: [ref 1, ref 2]. Despite being called a 'unit', it does not have a universal value; the value of the Weisskopf unit depends on the mass number of the nucleus in question and which transition the nucleus is undergoing ($E\lambda$ or $M\lambda$). The references contain expressions for the value of a Weisskopf unit in terms of $A$ ...


5

In nuclear Physics estimates can be made using the shell model of the nucleus of the gamma ray transition rates in excited nucleii and such estimates are named after Victor Weisskopf. A measured rate is compared with the Weisskopf estimate and the ratio is said to be in Weisskopf units (W.u.).


1

Actually, a nucleus has no electrons at all. The number of protons ($Z$) was never the same as the number of electrons (zero). Of course, most nuclei are part of complete atoms. If a nucleus that is part of an atom decays by alpha emission, it will typically also lose two electrons. That's not usually mentioned because it's not part of the radioactive decay ...


7

The popular press's description of this experiment is wildly wrong. It's hard to tell whether they just got it completely wrong on their own, or Scheck got it wrong and they're accurately describing what he said, or if it's some combination of the two. Scheck is a co-author but not the first author, and none of the ridiculous things they represent him as ...


55

To be honest, much of this feels like very irresponsible journalism, partly on the part of the BBC and very much so on the part of Science alert. If you're looking for an accessible resource to what the paper does, the cover piece on APS Physics and the phys.org piece are much more sedate and, I think, much more commensurate with what's actually reported ...


29

The articles are a little on the hysterical side, but I think they are just saying that violation of CP-symmetry means there must be violation of T-symmetry. T-symmetry means that physical laws are unchanged if we reverse the direction time flows. Classical theories obey T-symmetry, and it seems intuitively obvious that quantum mechanics would as well. But ...


2

Yes, the radiation both in Hiroshima and Nagasaki is very low, well, operationally non-existent. The radiation levels match the world average background radiation of 0.87 millisieverts per year. The bombs were optimized to have the maximum destructive power. That included a rather high altitude. The Hiroshima and Nagasaki bombs exploded in 580 and 500 ...


0

From the Wikipedia article on the Trinity site where the first atomic bomb was tested. This is the same type of bomb as used over Nagasaki. More than seventy years after the test, residual radiation at the site is about ten times higher than normal background radiation in the area. The amount of radioactive exposure received during a one-hour visit ...


1

Actually, the critical mass is not affected by external gravitational fields. Gravity does not effect nuclear reactions. And, for the most part, gravity does not effect even chemical reactions either. Intermolecular forces (electromagnetic in nature) are vastly stronger than gravitational forces. Rather, the critical mass has to do with the effective ...


0

The mass is the same to a very good approximation: gravity is absurdly weak compared to the factors that influence this. One way that it might influence things was if the shape of the mass was significantly macroscopically distorted by gravity (ie a sphere might become flattened and you might therefore need slightly more mass). However metals are quite ...


0

I'm failing to understand why helium is anomalous on the binding energy curve. I know it includes concepts such as pairing, shell correction, and the liquid drop model. However, these were simply stated by my professor. He didn't explain any of them. The mass of a nucleus is always less than the sum of the individual masses of the nucleons(n,p..) which ...


4

The number of atoms in your radioactive sample falls exponentially with time, so we get something like: $$ N = N_0 e^{-t/\tau} $$ where $\tau$ is a characteristic constant decay time called the mean lifetime. The half life is then defined by: $$ \frac{1}{2} = e^{-t_{1/2}/\tau} $$ or: $$ t_{1/2} = \tau\ln 2 $$ By this reasoning a $3/4$ life would be ...


0

Half life is a common physics misconception. Half life does not actually mean the cencentration of a substance being reduced to half by radio active disintegration. But it actially means the activity of the substance being reduced to half. For example if a sample of wood is found have uranium content with 5000 disintegration per second. And the half life is ...



Top 50 recent answers are included