# Tag Info

14

Actually, all the atoms are identical. The time at which it is observed to decay is not an intrinsic property of a given atom, but rather an effect of quantum mechanics. For any given time bin, there is some finite amplitude for a transition to a decayed state, which in turn corresponds to a finite probability, since the particle(s) emitted will escape from ...

10

The reason why alpha particles heavily dominate as the proton-neutron mix most likely to be emitted from most (not all!) radioactive components is the extreme stability of this particular combination. That same stability is also why helium dominates after hydrogen as the most common element in the universe, and why other higher elements had to be forged in ...

9

To pretty much everything you stated in your question, "no". That convection requires a medium is not the main difference, it is simply the most obvious aspect of what is a fundamentally different mechanism for transfering energy. Convection is the transfer of energy by movement of a medium, whereas radiation is the transfer of energy by, well, thermal ...

9

You can get an upper limit by simply treating the case of The radioisotope structured as a point source. The whole dose still present, corrected for half-life. No shielding. A simple $\frac{\text{presented area}}{4\pi r^2}$ for the acceptance (in this case the radiation is emitted in all directions, acceptance represents the fraction that hits the target, ...

8

Ultra-high energy cosmic rays all come from a very, very long way away (anything with the power to create them nearby would constitute a danger to life as we know it). I think the preferred mechanism these days is dynamic acceleration in the jets formed by active galactic nuclei, but don't quote me. Anyway, ultra-relativistic though they are, that means ...

8

The simple answer is no, though as usual in Physics things are a bit more complicated than that. There are several ways in which radionucleotides decay: alpha decay, beta decay, gamma decay, and fission. These are all mediated by the weak and strong nuclear forces, though the electromagnetic force plays some part in alpha decay and nuclear fission. There is ...

7

Short answer: no. Longer answer: No, excepting neutrinos none of the products of radioactive decay has the penetrating power to pass through the atmosphere, and neutrino detection is not something we can do from satellites. To elaborate, the immediate products of radioactive decay are (some set of, depending on the decay in question) fission fragments, ...

7

It is not a matter of "falling in": all s orbitals have non-trivial probability densities at the center. It is about energy balance in the nucleus. Kr-83 is a lower energy configuration than Rb-83 by enough to make up for the neutrino and the gamma(s). Evidently Kr-85 is not a sufficiently lower energy state than Rb-85.

7

It may have something to do with the growth rate of sunflowers. During peak growing times sunflowers can grow inches in a single day, which likely results in them drawing more water out of the ground, allowing them to concentrate the radioactive materials through deposition in the plant matter at a faster rate than other plant organisms. I would suspect ...

6

The color of a surface doesn't reliably indicate the emissivity at non-visible wavelengths. The color in the visible spectrum is more of a side effect than anything. Most thermal radiation around body temperature or room temperature happens in the infrared region, not the visible, and that's not reliably indicated by visible color: The transparent ...

6

I hear that the WC-135 Constant Phoenix has recently been deployed to monitor radiation in the air around Japan. The Vela satellites which were operational until 1984, and currently the DSP satellites, are intended to give immediate reports of nuclear bomb detonation and ICBM launches. The Vela satellites included gamma ray detectors, which accidentally ...

6

Indeed, it's likely that the rice came from Japan, and if it did, it's pretty likely it came from the Fukushima region which is famous across Japan and beyond for its rice - and other products. However, it could have been harvested before the tsunami. But as discussed here, Are these radioactive particle matter and air emmissions dangerous, 2000KM from ...

6

There's certainly no problem being out in the Sun during an eclipse: There's nothing being emitted then that's not being emitted at other times. The danger is just that the relative darkness near totality may make it seem safe to look at the Sun, even when it's not. But as long as you don't look directly at the Sun, you're fine. During the time when an ...

6

Thermal neutrons capture on hydrogen and carbon with reasonable (i.e. not large, but significant) cross-sections (this is the delayed event detection methods of most organic liquid scintillator anti-neutrino detectors--i.e the one that don't dope their scintillator with Gadolinium). So though a "cloud"--meaning a localized diffuse gas--of neutrons can ...

6

Pasted text of the letter in English - the link also contains the original typed German letter. Open letter to the group of radioactive people at the Gauverein meeting in Tübingen. Zürich, Dec. 4, 1930 Physics Institute of the ETH Gloriastrasse Zürich Dear Radioactive Ladies and Gentlemen, As the bearer of these ...

6

As @dmckee says the problem is complicated. It is complicated because it is not a solution of a potential describing one force, but a balance between electromagnetic forces and the strong force that is keeping the quarks within the nucleons. (In the nucleus the strong force is like a type of Van der waals potential, a higher order interaction, overflowing ...

5

First, just recall that one becquerel is one radioactive decay per second. If your spinach has had 1-3 Bq/kg, then you should just completely forget about any threats - there is no danger whatsoever. The safe limits are 100-500 Bq/kg, see e.g. http://www.anarchyjapan.com/2011/04/for-now-no-changes-in-food-radiation-safety-levels/ which are 100+ times ...

5

Yes. Have a look: “The ‘Reifenschweiler effect’ is the observation that the beta-decay of tritium half-life 12.5 years is delayed reversibly by about 25-30% when the isotope is absorbed in 15 nm titanium-clusters in a temperature window in between 160-275 C. Remarkably at 360 C the original radioactivity reappears. The effect is absent in bulk metal. ...

5

There is nothing magical about lead for this purpose. The driving factor is the number of electrons per unit volume, which reduces (to a first approximation) to the mass density. You get very good (better than lead) shielding performance from gold, tungsten, mercury, etc; and quite reasonable performance from iron or copper. Question for the student: why ...

5

The bad news: Space radiation is much harsher compared to boring gamma rays from our primitive nuclear reactors. Space radiation has much higher energy levels, and you cannot completely shield it, even with 10 meters of lead (which is in fact not very effective for neutrons). The good news is that an individual gamma photon, for example, usually would not ...

5

I agree with qftme's answer for the case of massive decay products. By energy conservation alone, $\gamma \rightarrow e^+e^-$ should be allowed, but momentum conservation forbids it (as well as the opposite case, $e^+e^-$ annihilation). It is only allowed if you have some other particle involved to take care of the photon momentum. In case of a massless ...

5

The power going into an empty microwave will accumulate as an EM field inside the cavity up to the point where the input power equals the rate of leakage. Energy inside the cavity will leak out through the little holes in the metal screen in the door, through ohmic losses in the sided of the microwave, feedback into the power source, etc. The rate of ...

5

It's actually a bit unlikely for the following reason: Dark matter is observed to not 'clump' as effectively as ordinary matter. This is due to the fact that ordinary matter feels drag due to interactions with the intergalactic dust and loses angular momentum to radiation, causing it to fall inward toward the galactic core, and to fall toward a central ...

5

If the neutron decayed to a two body state (any two body state) the energy spectrum of the products in the neutrons rest frame would be single valued (this is required by the conservation of energy and momentum). It is not. Instead the electron energy spectrum is a continuum that runs from that roughly the two-body limit down to as near zero as our ...

5

Neutron decays into a proton, electron and electron anti-neutrino. Not only electric charge but also (electronic) lepton number has to be conserved (I'm not very sure in this statement). In short: you start and have to finish with 1 matter particle (anti-matter counts as -1). Mean time $\neq$ half-life. More on wikipedia. Physically there is no difference ...

5

Heat is not caused by thermal energy being radiated from particles due to their energy heat is the ramdomized (i.e. neglecting bulk flows) energy of motion in any material (including, for instance, photon gases). Any vacuum that we can make or have access too includes a small amount of matter, and the temperature of that stuff can be measured. Not ...

5

The neutron is made of two down quarks and an up quark; the proton of two up quarks and a down quark. This leads to two effects that differentiate their masses. One is that the up and down quark themselves have different masses. The other is that the proton is charged, and so quantum corrections involving virtual photons affect its mass. The details are ...

5

When you have a three-level system, the laser transition is between the ground and first excited levels (see figure). In this scheme, it is rather challenging to get population inversion because all atoms tend to stay on the lowest level. With a four-level scheme, you have an extra level so that the laser transition does not end in a ground state. Thus, if ...

5

Hint :You have masses (from parent nuclei mass you can get mass of daughter nuclei by subtracting mass of $\alpha$ particle , and the Q value ie. the energy that gets liberated and $931.5 \ MeV \approx 1 amu$ and so you can see the excess mass is easily negligible ie $<0.1amu$ $^{212}Po \rightarrow \ ^{208}D + \ ^4\alpha$ momentum is zero ...

5

It is a matter of confusing terminology , at the present times when so much differentiation has happened in physics related scientific disciplines. Radiation was a general terminology assigned to transfer of energy radially, to start with with waves: acoustic waves, waves in water. Then came Maxwell's equations and the discovery of electromagnetic waves, ...

Only top voted, non community-wiki answers of a minimum length are eligible