swansont

here:

Wouldn't the "ridiculously" cancel out, then? c2 is almost 1 GeV, and electron transitions are of order 1 eV, so the mass change is (very) roughly a part in 109

(volume turned down by me) Gammas are photons from nuclear interactions. They are waves if you look at wave properties, they are particles if you look at particle properties. Gammas can create electron-positron pairs if of sufficient energy; they can cause the photoelectric effect and undergo Compton scattering. Since we were talking about a quantum of energy produced in an interaction, there is really nothing terribly wrong with saying "gamma particle," quasi-redundant though that might be. Certainly nothing so egregious as to elicit your response.

Generally speaking, the only people selling science on the web are the crackpots. It's quantum snake oil.

The Schwarzschild radius for a black hole of mass M is given by rs=2GM/c2, where G is Newton's gravitation constant.

The point between the earth and the moon where the force is zero is a point of unstable equilibrium. Sooner or later an object there will be perturbed, and start feeling a force. It is one of the five Lagrange points, called L1. There are three points along that line, and all are unstable. But there are two more that are stable - check the link for more details, and Google on 'Lagrange points' edit to add: Note that there are Lagrange points for both the earth-moon and earth-sun systems. The original question was about the earth-moon system; this link is to an explanation that shows the earth-sun points. But the concept is the same.

The meter was originally defined in terms of the distance from the equator to the north pole, being 10,000km. From that they defined the standard meter in terms of an iridium-platinum bar. The number was a little off because they had not accounted for the oblateness of the earth, so the actual measured distance is not quite 10,000 km. I've never read where c enters into this, or if without this discrepancy c would be a round 3e8. Today, of course, c is defined and the meter is a certain number of wavelengths of light from a specified atomic transition. But that's not where it originated, AFAIK. No, objects with zero rest mass travel at c. Travelling at other speeds means there is a mass term - if v>c then this term is imaginary. The superluminal experiment to which I think you are referring was a case of anomalous dispersion and did not in any way constitute a case of a photon being accelerated to move faster than c. It was basically reshaping a pulse of light (many photons) so that the peak was further ahead afterwards than in the original pulse.

On the high end you are limited by how much energy is available to you, and the size of your sample. At the low end, it's zero, as the Adm stated. people have come close - below a nanoKelvin - but the limit is unattainable.

Technically, no. Gammas come from nuclear interactions. Betas being slowed down will emit Bremsstrahlung X-rays. Doesn't need to be wax - just about anything will do. Slamming energetic electrons into metal is a standard way to make X-rays.

Am-241 is an alpha source. Betas generally won't kick neutrons out of a nucleus edit:fix quote tag

It wouldn't. I imagine the concept of requiring infinite energy to travel at c for massive objects, and only massless objects - which have purely kinetic energy - travelling at c, can somehow be tortured and mangled into that statement. But it's not something I would care to watch.

My lab can be found in links here (scroll to the bottom) and here

Shouldn't 7 alpha decays be [math]^{206}_{78}\mbox{Pt}[/math] ? I realize this may just be an exercise in writing out the steps' date=' but there's no way that this result is physically real. [math']^{214}_{82}\mbox{Pb}[/math] only beta decays. You have to include that, along with some other (unlikely) branching, to arrive at [math]^{206}_{80}\mbox{Hg}[/math]

You've violated spin and lepton number conservation. [math]\left.^{11}_{4} \mbox{Be}\right. \rightarrow ^{0}_{-1}\beta + \left.^{11}_{5} \mbox{B}\right + ^{0}_{0}\bar{\nu}[/math]

Why not?

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To clarify: it's not that the alphas are "weak," i.e. low energy - it's that they deposit their energy in a very small distance of travel. Internal exposure is really bad. External exposure is no big deal. Any you're right- gammas are a "whole body" dose. But I still don't recommend eating any.

Yes, but that was fairly early on - certainly not within man's time on earth, and thought to be well before that. I think the composition of the moon indicates that it was after the chemical differentiation of the core, mantle and crust, since there's not a lot of iron on the moon.