New answers tagged

1

What I cannot understand and what my textbook does not mention is where the 0.61 part comes from. How is this derived and what is the significance of 0.61? In your equation if you take nuclear diameter $D=2R$ Then your relation looks like $ D\sin\theta = 1.22 \lambda$ If one wishes to resolve two objects separated by a distance $D$, one can use ...


0

Absorbed dose for a given tissue or body organ can not be easily calculated using simple equations, even though it is simply defined as "amount of mean energy imparted per unit mass" , because it depends on so many other factors. Absorbed dose for a given material can be measured using devices called as Dosimeters such as (Calorimeters, ion-chambers, or ...


5

Electron capture Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission ...


8

The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays ...


1

RAD, Radiation Absorbed Dose 1 RAD = 100 ergs absorbed per gram. In my opinion there is no simple formula to calculate the dose in a very straight forward manner. To calculate the dose you have to calculate the energy deposition in a mass. For this you need range vs energy deposition curves. In the case of heavy particles they show a so called Bragg peak i....


0

Wikipedia's article on radioactive decay gives the equation that describes a two-step decay chain, together with the corresponding solutions. In your case, you'll have $$\begin{align} N_{\text{Kr}}(t) &= N_{\text{Kr}}(0)e^{-\lambda_{\text{Kr}}t} & N_{\text{Rb}}(t) &= N_{\text{Kr}}(0)\frac{\lambda_{\text{Kr}}}{\lambda_{\text{Rb}} - \lambda_{\text{...


0

I would suggest "Introduction to Nuclear Engineering" by John Lamarsh. The text is aimed at juniors and seniors with no previous training in nuclear engineering. It starts at quite a low level, even introducing the fundamental particles, concepts of particle wavelength related to particle energy etc. It is mainly concerned with nuclear reactors as that is ...


2

There is not, because the combined transformation $CPT$ is a symmetry of all Lorentz-invariant systems. The $P$-violating decay distribution observed by Wu et al. is also a $C$-violating distribution, because polarized anti-cobalt would have had the opposite sign of asymmetry. (However no one has ever made, or probably will ever make, polarized anti-cobalt,...


3

The trick is to convert to radians. You're mixing radians and degrees, which I think makes your error too big by $180/\pi$.


2

There's some missing background information: all even-even nuclei have ground state spin and parity $J^P = 0^+$. Suppose a nuclear excited state with $J^P = 0^+$ were to decay to the ground state by emitting a single photon. The lowest angular momentum that can be carried by a photon is $\hbar$, in dipole radiation. (The parity of the EM radiation ...


2

the typical spontaneous emission time scale in atomic physics is on the order of 10^−6 s. In contrast, in nuclear physics, many radioactive nucleus have a half-time of 10^6 years or even more. I beg to humbly disagree with your picture of generalization of atomic and nuclear time scale of events and putting up a contrast/relation with the atomic and ...


2

The slow nuclear transitions have a potential barrier so they proceed by tunnelling and the rate is supressed by a factor of $e^{-E/E_0}$, where $E$ is the barrier height and $E_0$ is some characteristic energy. Your golden rule calculation is giving you the rate in the absence of a barrier. Potential barriers are rare in atomic physics, but frequent in ...


8

To be honest, the paper listed as reference 12 in the OP's quote (1) gives as good an answer to most of these questions as you can hope to get: For example, taking $\omega(E)$ as a Lorentzian function for all E yields the well-known exponential decay at all times. However, in real physical systems, $\omega(E)$ must always have a lower limit, which ...


1

All mesons are unstable and most have quite short half-lives even by particle physics standards, so they are only to be found (as on-shell particles1) in the presence of energetic events. And if you look closely enough the process that creates them is essentially2 always one of banging two particles together pretty hard. 1 That is, I am excluding virtual ...


3

There is something famously called "Feynman's famous formula", which comes up in QFT calculations, which I imagine must be the second FFF referred to in Welton's account. It reads: $$\frac1{a_1 a_2 \ldots a_n} = \int_{x \in \Delta^{n-1}} \frac1{(\sum_{i=1}^n a_i x_i)^n} d\sigma$$ where $\Delta^{n-1}$ denotes the simplex $\{x = (x_1, \ldots, x_n) \in \...


-8

Since no nuclear bomb has been used since since Nagasaki we can and indeed do only infer "applications" if you are talking about dropping a nuclear weapon on a human population. These estimates based upon the criteria you have given appear very effective as no theronuclear weapon has been detonated on a population center since Nagasaki, Japan in 1945. I ...


24

The so-called TNT equivalent of a nuclear weapon is an unambiguous way of quantifying how much energy is released by the nuclear weapon. There's nothing 'wrong' about it. The only caveat is that the damage caused by, say, Little Boy versus 15 kilotons of TNT would not be identical despite having an equivalent yield (for various practical reasons). ...


0

It turns out that this structure is the a doppler broadened line coming from the $^{10}$B(n,$\alpha$) reaction which populates the 477.6 keV excited state in $^7$Li.


1

is driven by the electromagnetic force (there is a γ in the Feynman's diagram). you are putting the cart in front of the horse. Interactions are defined by their strength. Stronger, i.e. most probable is the strong interaction and the crossections, i.e. probability distributions , are much larger than the electromagnetic interaction, which is larger than ...


1

The spectrum of emitted rays is defined by the energy output of the reactions happening in the device and its surroundings. Basically, after an exothermic nuclear reaction, the produced particles smash into surrounding material and bounce for some time until their energy is dispersed. This energy heats up the medium which begins to emit black-body radiation ...


4

The earliest reference I've been able to find on the half-life of 235U is in The Uranium Half-Lives: A Critical Review, by Norman Holden, which reviews various early studies of each of the common isotopes of uranium (232U, 233U, 234U, 235U, 236U, and 238U). The earliest study he cites is Nier (1939) (A. 0. Nier, The isotopic constitution of uranium and the ...


1

A neutron has baryon number = 1, while the anti-neutron has baryon number = -1. Physics Guy has much the same here with quarks, that works as well. In the language of CPT the charge operator reverses the charge of the quarks, so the two up quarks with charge $-1/3$ is flipped to $1/3$ and the down quark from $2/3$ to $-2/3$. I gave this question a 1-vote, ...


4

Every particle has (or can have) an antiparticle. Sometimes, it is even his own antiparticle. An antiparticle $D'$ is (easily said) defined as a particle $D$ after a CPT-transformation. CPT-Symmetry is believed to be a fundamental concept of physical nature. A CPT-transformation is a complete changing of observables of a particle. The C stands for ...


-2

The density is not high enough in Jupiter or Saturn to achieve a sustainable environment for the fusion of hydrogen and helium.


3

Well yes, maybe, but they are called planets. So fission in stars? No, but maybe in planets. I do not know what the status of this is, but the core of the Earth is heated by weak and maybe strong nuclear processes. The standard model is that weak nuclear decay. The major heat-producing isotopes within Earth are potassium-40, uranium-238, uranium-235, and ...


0

No. Basically because stars, by definition, are fusion reactions. It's feasible for a star to contain Uranium - the Earth, for example, contains Uranium, and could (in fairly extreme circumstances) collect enough matter to become a gas giant, continue to collect matter and then start fusing to become a star. But even then, you'd just have a star containing ...


0

To perhaps give an answer in a slightly different way, it's complicated. One must keep in mind that the decomposition of the angular momentum into spin and orbital components is model dependent. In the language of the renormalization group, it is scheme and scale dependent. If you probe a hadronic system at a given length scale (or, equivalently, energy) ...


1

Indeed, a nuclear bomb works a bit differently in a vacuum than in the atmosphere. If you want to generate momentum, an atmosphere or some material with a low boiling point (e.g. ice) is probably better than bare rocks in vacuum, because all the heat will be converted directly into gas with a high momentum, without "wasting" energy on heating and evaporating ...


0

Quarks as elementary particles are quantum mechanical entities. The same is true of electrons and protons. The hydrogen atom has the simplest quantum mechanical solution of how quantum mechanical particles are bound by attractive forces. This solution is a wave function, and it tells us that the electron is in a quantum mechanical probability locus around ...



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