# Tag Info

0

Actually, I think that the atoms would undergo "coherent radioactive decay", a process analogous to "coherent elastic scattering" (which is well to occur in some neutron scattering experiments, and played an important role in speculation on relic neutrino detection in the 80's). For $N$ radioactive atoms with a lifetime $T$, one normally expects that the ...

1

First off, Strontium accumulates in the bones replacing Calcium, so you can't easily get rid of it. Second, the most radioactive natural food we eat is the banana, at about 130 becquerels per Kg. So that 1L of "hot" water really is quite nasty if you drink it and a significant portion of the Sr90 ends up in your bones.

2

You're right that in the context of radioactivity, antineutrinos are pretty much only released when a neutron turns into a proton, ${}_0^1n\to {}_1^1p+{}_{-1}^{\;0}e+\bar{\nu}$. They can also be consumed when a proton turns into a neutron and a positron, ${}_1^1p + \bar{\nu}\to {}_0^1n + {}_{1}^{0}e$. There are some other processes that involve ...

1

There are many types of nuclear decay, and many techniques for estimating half-lives. For beta decay of states in spherical nuclei, calculation of decay rates is a classic application of the (spherical) nuclear shell model. For gamma decay, there are generic estimates that are based on the energy and multipolarity of the transition. (The term to google on ...

3

Here is a table of isotopes versus lifetimes the color code of the lifetimes on the right hand column: Isotope half-lives. Note that the darker more stable isotope region departs from the line of protons (Z) = neutrons (N), as the element number Z becomes larger Modeling a nucleus is a many body problem and also a many forces problem. There exists ...

3

The transition probability per unit time of a nucleus from an initial state i to a final state f, representing the decayed system, is modeled by Fermi's Golden Rule: $$\lambda=T_{i\rightarrow f} = \frac{2\pi}{\hbar}\left|\left\langle i\left|H'\right|f\right\rangle\right|^2\rho$$ Where $T_{i\rightarrow f}$ is the transition probability from state $i$ to state ...

-1

As far as I'm aware, this is just a phenomenological constant. That is, you get a few people looking (well, not exactly "looking") at a few mols of X radioactive element and. After they have enough measurements, they will get said constant through boring data fitting. Worked in it myself! It does not have a clear dependance on anything (and we have ...

4

The constant is a function of the stability of the nucleus, and is experimentally determined for every isotope. In other words - every kind of nucleus has its own value of $\lambda$ and there is no way (that I know) to get an accurate value for it, other than measurement. But there are some nuclear physicists roaming who will put me out of my misery, I'm ...

0

Yes. After beta decay the atom has charge +1. Electron has energy about 1MeV but due to scattering on the another atoms lose energy and stop.

1

If we ignore neutrinos, which are weakly interacting, radioactivity is still classified as alpha beta and gamma. The energies are of order MeV. Of these three, only gamma is neutral and has a chance to cross the atmosphere . Then one has to take into account the 1/r**2 diminution of the flux for the large distances . To localize a source another ...

2

Not possible in practice, even though neutrinos emitted by the Plutonium might be used in principle if we ever found a way of intercepting them with almost 100% efficiency. However, there is/was a scheme to use neutrino analysis to determine whether a reactor is being used to create Plutonium

0

So the atom is left with two more electrons than protons and therefore has a negative net charge. This repels the two electrons, so they leave, and fast. And now the atom is neutral.

9

It's not true that the atom is electrically neutral afterward. If you have a single atom, isolated in a vacuum, and all that is emitted is an alpha particle, then as you say, it has a net charge of -2e. In reality, alpha decay is a violent process that is likely to knock out some electrons as well. Furthermore, if the atom is in a solid, then electrons are ...

2

You're right, directly after the emission it will have charge -2e if it was neutral before, i.e. be an ion. But within a gas or liquid electrons are very easily exchanged and ejected. This is of course at a much lower energy energy scale than the nuclear emission and therefore less noticeable.

3

A decay destroys the electromagnetic wave function of the atom, the one that generates the energy levels which keep the electrons bound to it. The new nucleus, after the deacy, will have a new potential whose solutions will have binding levels for n-2 electrons. The two extra will be left in the lattice ( or in the gas) free to join up in the energy ...

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