# Tag Info

41

If protons decay, then what you say is true: all atomic nuclei are indeed unstable, and a so-called "stable" nucleus simply has too long a half-life for its decay to be observed. The most tightly bound nucleus is $^{62}$Ni, with a binding energy per nucleon of 8.79 MeV [source], which is less than 1% of the mass of a nucleon. On the other hand, the decay of ...

23

We are never 100% certain of anything. The scientific method falsifies wrong theories, but it does not verify those we colloquially call "correct" or "true" If we tomorrow detect a normal oxygen atom decaying, we'll have to devise new theories to explain it. But we don't expect the things we call stable to ever decay (that's why they're called stable). We ...

8

In regards to a single element, this actually happened. Natural Bismut is about 100% 209Bi, which shows no obvious indication of being radioactive. But it turned out that all existing bismuth is radioactive. The isotope 209 had been suspected to be unstable before, but that was experimentally verified as recently as 2003, finding a half life of 1.9×1019 ...

8

As John Rennie answered it very clearly, I would like to add some more details too. See, around early 1900 the idea of atoms was floating around the scientists' heads. At first everything was theory, but these things happened: You certainly heard of Joseph Thomson's cathode rays. Well, he actually calculated the ratio Q/m of atoms. (You can search any of ...

6

According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy. There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc. Only the first one isn't described in Bohr theory - it's ...

6

The answer is yes, the atom does absorb radiation that does not exactly match the transistion frequency. This is due to the Doppler effect that everyone knows from an ambulance with siren driving by. The frequency you hear is higher if the ambulance moves towards you and lower if it drives away from you. It's the same with the atom. If the atom moves (and ...

5

There is no such experiment, though there are lots of experiments where the number of electrons in an atom are measured as a side effect. We know atoms are electrically neutral so there must be equal numbers of electrons and protons. We know successive elements in the periodic table are built up by incrementing the number of protons, so we know how many ...

5

My answer has a lot of overlap with the earlier answer by NoEigenvalue, but grew too long for a comment. The orbitals are segregated by the amount of angular momentum that they carry. No angular momentum, $\ell=0$, are called $s$-orbitals, and are spherically symmetric. The first two examples in your figure are $s$-orbitals with $n=1$ and $n=2$. Angular ...

4

The graph shows the probability of finding the electron between the distances $r$ and $r + dr$. This probability is given by: $$P = \psi^* \psi dV$$ where $dV$ is the volume element: $$dV = 4\pi r^2 dr$$ So we get the probability: $$P(r,r+dr) = \psi^* \psi 4\pi r^2 dr$$ and therefore when $r = 0$ the probability $P = 0$. It isn't that the ...

4

Finally they point out that the probability density for the electron can be regarded as a charge density, a stationary cloud of negative charge. But, this is wrong, it only a "probability density" in the sense that this is the probability to find a particle at a given location for a given associated (hypothetical) large number of identical systems ...

3

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about ...

3

Just wanted to add that it's not totally true that the drawn orbitals are "the regions where an electron can be found". But my answer grew and grew... Let's take a neutral boron atom where it has filled 1s and 2s shells and one electron in the 2p shell. Suppose it is floating in space, far away from messy interactions with other things. So, you wonder, ...

3

The orbitals are only spherically symmetric for $S$-states, for which the angular momentum number $l$ is zero. So in you picture the first two orbitals are the $s=1$ and the $s=2$ state, while the other three pictures correspond to the $n=2,l=1$ with the three different possibilities of $m_l$. There are several ways to see that the wavefunctions with $l=0$ ...

3

The ionisation energy of a nitrogen atom is about 14.5eV but in $^{14}$C beta decay the electron is emitted with an energy of 156keV, which is far higher than the ionisation energy. So beta decay will normally produce an N$^+$ ion and a free electron.

3

No! In chemistry this is called "Radical Ions". some of the radicals can exist, but it is very hard to find them in the nature because they interact very fast. So the question is how can you understand that a radical exist or not? well you have to know "Physical Chemistry". the Radical that You mentioned can not exist, because of high density of charge ...

3

I would like to answer your second and third questions first. Although there exist good approximations of the force due to strong nuclear force, a much easier option would be to compare the binding energy with the potential energy created due to electrostatic repulsion. To make an estimation of the potential energy, I assume the radius of Helium nucleus to ...

3

Why is energy released when an atom decays into another atom, when no energy is added? Atoms/nuclei are already created when we study them and organize them into the periodic table of elements. At the level of nucleons and elementary particles in general, special relativity holds. When we look at the periodic table of elements and count the number of ...

3

Radioactive decay changes an atom from one that has higher energy inside its nucleus into one with lower energy. The change of energy of the nucleus is given to the particles that are created. The energy released by radioactive decay may either be carried away by a gamma ray electromagnetic radiation (a type of light), a beta particle or an alpha particle. ...

2

For some reason I get 28.296 MeV for binding energy of $^4He$. Coulomb potential energy (for a charged sphere) can be written as $E=\frac{3}{5}(\frac{1}{4\pi\epsilon_0})\frac{Q^2}{R}$. This term is actually plugged into a liquid drop model formula (the Coulomb term $-a_C\frac{Z(Z-1)}{A^{1/3}}$) and from both theory and experiment give you a similar value ...

2

It is a very interesting question. First point is that even with 1 meV of energy the electron will have plenty of states. I suppose the issue is what exactly is a zero energy electron. Hotop et al. have done very nice experiments in electron scattering where they control the electron energy to below 1 meV by laser photoionization. They can then observe the ...

2

This is a tough one. I'm going to take the liberty of rewording the question, and then use analogies to hopefully give some kind of answer that gets us part of the way there with this arguably obscure but worthy issue. How did the electron change from simply flying past the proton to adopting a quantised orbit around it? Because it was attracted towards ...

2

When you have only one electron then $\Delta S=0$ makes intuitive sense: you can change the angular momentum $l$ of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change $s$. Would it be possible to change $s$ then you could ...

2

Your expression "the angular momentum is $m_j \hbar$" (where $\hbar = h/2 \pi$) is incorrect. This quantity is the projection of the angular momentum on the $z$-axis; it represents the direction of the spin. This is why it corresponds to the $m$ quantum number, not the $\ell$ quantum number in the $| n \ell m \rangle$ basis. Which angular momentum to use ...

1

Comments to the question (v2): The fact that the TISE is invariant under a symmetry group $G$ (in this case the Lie group $G=SO(3)$ of 3D rotations) does not imply that the orbital/wave-function solutions $\psi$ must be $G$-invariant as well. (Think e.g. on spontaneous symmetry breaking where the governing laws of a physical system are invariant under a ...

1

Yeah, they can have a representation as a wave. What you have to realize is that we have absolutely no intuition about the world at particle scales. Here, in the macroscopic world, we easily distinguish thing as particles or waves. For example, sound and the motion of a string are clearly waves to us, while a basketball and a car are clearly particles. Non ...

1

Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ...

1

What makes you think that the question has an answer? The Bohr model has limited validity and this was realized from the start. In essence, you're describing a transition from an unbound state of the electron (with positive total energy) to a bound state (with negative total energy). This cannot happen all by itself, as the extra energy needs to go ...

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