I have seen multiple claims that the neutron is its own antiparticle. However surely the antiparticle of the neutron would have a baryon number of -1 because it would consist of three antiquarks, each of which would have a baryon number of -⅓. I could be mistaken but I'm assuming that baryon number is flipped when taking antiparticles.
107449
The anti-particle corresponding to a proton or an electron is a particle with an equal mass, but an opposite charge. So what is the anti-particle corresponding to a neutron (which does not possess a charge)? And if it is just another neutron, will its collision with the original neutron be as destructive as the collision of a proton with an anti-proton or an electron with an anti-electron?
24599
A nucleus consists of protons and neutrons. Both are extremely heavy compared to electrons. Then how come they are contained within an extremely tiny space? And why does the atom consist of 99.999% empty space?  I do not understand the mathematics regarding this one bit. Please give me an explanation of this phenomenon in words as far as possible :)
79992
In the book of The First Three Minutes by Weinberg, on pages 106-107, it is stated that  SECOND FRAME. The temperature of the universe is 30,000 million degrees Kelvin [...] The nuclear particle balance has conse- quently shifted to 38 per cent neutrons and 62 per cent protons. [...] THIRD FRAME. The temperature of the universe is 10,000 million degrees Kelvin. [...] The decreasing temperature has now allowed the proton-neutron balance to shift to 24 per cent neutrons and 76 per cent protons.  What is the reason for this balance shift between neutrons and protons? And what determines the rate of change of the neutron/proton ratio?
71879
The mass of nucleons is bigger than the sum of the masses of its constituent quarks. I understand that it's because you have to take into account the binding energy of these quarks.   What I don't understand is this: In a nuclear physics course I was tought that the mass of the nucleai is less than the sum of the masses of its constituent nucleons because the binding energy between them is negative in order to make it stable.   So, if the binding energy of nucleons is negative because its suppose to be an atractive force that keeps nucleons stable, why does the binding energy of quarks add to the mass of nucleons instead of reducing it if it's also an atractive force keeping quarks stable?  I understand that in the quark regime we have to deal with quantum chromodynamics and the concept of force and energy are not that simple but I was expecting the binding energy of any atractive interaction to be negative anyway...
81873
A deuterium nucleus is composed of a proton and a neutron. Both have spin $\tfrac12$ so I would expect the deuterium to have two possible spins: $1$ for the triplet and $0$ for the singlet. But apparently deuterium always has spin $1$ and the spin $0$ state doesn't exist. Why?
18715
Below is experimental data for the ratio $$R=\frac{\sigma(e^+e^-\rightarrow hadrons)}{\sigma(e^+e^-\rightarrow\mu^+\mu^-)}$$ as a function of the centre of mass energy $\sqrt s$. I am interested in the peak at around $100GeV$ which corresponds to the resonance of the $Z^0$ boson. There are two ways of looking at this that I have in mind:  We can effectively ignore the electromagnetic process around this energy, and so each process should have it's own Breit-Wigner peak centred on the mass energy of the $Z^0$. The ratio of these is just a constant. Thinking in terms of Feynman rules and again ignoring the electromagnetic process around this region, each process has the same propagator and vertex factors (roughly - ignore quark mixing) and there are some extra factors due to different quarks being possible and colour degeneracy, but still the ratio should be a constant (one possible issue here is interference between Feynman diagrams which I have neglected?).  So my question is why does the peak exist in the data?
61394
Why do the four spin 3/2 $\Delta$ baryons have nearly identical masses and lifetimes despite their very different $u$ and $d$ quark compositions?
4529
I've always wondered why the number of protons in the Universe exactly matches the number of electrons. They are such different particles with totally different cross sections. So, first of all, is this true? And, secondly, how could it be that they have been produced in the same amount after the Big-Bang?
55997
This is something that has long bothered me, and I have asked a few physicists and chemists and never gotten a very satisfying answer. Why are particles charged? And I'm not asking (and this is the answer you often get) why some molecule might be negatively charged because the answer to that is simply is that there was one more negatively charged particle than positively charged particle. That doesn't answer my question, however, of why there are charged particles at all?  So, on a fundamental level, what makes one particle charged and another one neutral.  For instance, protons and neutrons have almost exactly the same mass, which I'm assuming that means they have very similar internal compositions, yet one is positively charged and the other neutral. What is going on that makes these particles different?  On a similar note, and I'd really like to know this, how is it that a proton and an electron have exactly equal and opposite charge, yet an electron is about 1000 times less massive than a proton?  I have particular trouble with this question because if the answer were something along the lines of "particle X which composes protons all have partial positive charges of n amount..." I would just ask why particle X has any charge and we'd be in the same predicament.  I'm sure there is a satisfactory answer to my question, I just can't find it and don't know what resources are reliable or detailed enough to be satisfactory.
3496
I heard that even a banana generates a minute quantity of antimatter. Does any know radioactive nuclear reaction produce antimatter along with alpha, beta and gamma radiation?
13794
I'm looking for explanation on  but unfortunately there's no Wikipedia entry or self-contained paper to start from. The statement of this effect is that:     "At leading order, multiple scattering only affect the momentum   distribution of the final particles, but not their total number. The   suppression at small p is compensated by an increase at larger p."   Can someone please explain why? I'm looking for a calculation based answer and not just qualitative discussion.
113784
With the confirmation that elements 113, 115, 117, and 118 are indeed fundamental elements that are now to be named on the periodic table, the next question is: what is the highest atomic number possible for an element? Feynman had a go at this years ago, and he derived (according to my limited understanding) that above element-137 (informally denoted 'Feynmanium') the electrons in the nearest orbit about the nucleus would be traveling at a velocity greater than the speed of light. (Note I am considering the Bohr model of the atom in this question.) I wished to demonstrate to my friends why this is and I came up with the following explanation.  According to Bohr's quantum condition, the angular momentum of an electron about the nucleus is given by $$ L=m_evr_n=n\hslash \implies r_n=\frac{n\hslash}{m_ev},$$ where $m_e$ is the mass of the electron, $v$ its velocity, $n$ an integer, and $r_n$ the radius of the $n$th possible orbit. Because we are concerned with the nearest orbit to the nucleus, $n=1$; and thus $$r=\frac{\hslash}{m_ev}.$$   Now, according to Coulomb's law, if the electron orbits about the nucleus the centripetal motion can be described by $$\frac{Ze^2}{4\pi\epsilon_or}=m_ev^2, $$ where $Z$ denotes the number of protons in the nucleus (the atomic number), and $e$ the elementary charge. Solving for $Z$ and substituting in $r$ from above yields $$Z=\frac{4\pi\epsilon_o \hslash v}{e^2}. $$ But what is $v$? Well, the maximum velocity an electron could ever have is the speed of light, and we wish to find the atomic number associated with an orbiting electron traveling at this speed, so we set $v=c$ and obtain our final result of $$Z=\frac{4\pi\epsilon_o\hslash c}{e^2}\approx 137.521,$$ which implies that for $Z&gt;137$, the electrons at a position of $n=1$ in the Bohr model would have a velocity $&gt;c$; and thus the highest atomic number achievable on the periodic table is 137.  Again, I just want to make sure this is a correct method for deriving element-137 before presenting. Perhaps one could explain how relativity plays a role here. I know Feynman used the Dirac equation to get this result...so could anyone (subsequently of course) expatiate on this in a simplistic manner? Thanks!
37015
It seems to me that we could change all the current spin values of particles by multiplying them by two. Then we could describe Bosons as even spin particles and Fermions as odd spin particles. Is there some consequence or reason why we can't simply change 1/2 as the smallest unit of spin into 1?  It just seems prettier this way. In fact, there's almost an interesting relation to even and odd functions in that if you switch any two bosons around the wave function stays the same while if you switch any two fermions around the wave function becomes negative.
32252
Free neutrons are known to undergo beta decay with a . Binding with other nucleons stabilizes the neutrons in an atomic nucleus, but only if the fraction of protons is high enough (at least a third or so). But what keeps a neutron star stable against beta decay? Apparently, this is extra pressure due to gravity in contrast to "negative pressure" of proton Coulomb repulsion in a nucleus but how do we know that this is enough to stabilize the degenerate neutronic fluid?  I am aware of a  but not really happy with the answers there. There is lot of dazzling details , but I am a looking for an answer suitable form a 8-year old with enhanced curiosity towards astrophysics.
101807
Neutron mass: 1.008664 u  Proton mass: 1.007276 u  Why the discrepancy?  On a related note, how does one go about measuring the mass of a neutron or proton, anyway?
31491
Title says it all: How can  and  both have $uds$ quark content? Doesn't this make them the same baryon?
105732
From the perspective of a matter-based observer does antimatter:   Increase in entropy (and therefore decrease in entropy in its own time)   OR   Decrease in entropy (and therefore increase in entropy in its own time)   Option 2 would seem to explain why we don't see much anti-matter (it all went into energy.)  Which is correct? Does antimatter increase or decrease in entropy over time?