Physics Asked on October 29, 2021
Generally in text books they say that when a electron goes from high energy state to a lower energy state it emits photons.
My question is, it is possible that a proton that goes from high energy state to a lower energy state emits photons too?
This answer addresses the title rather than the comment text, but all nuclei, regardless of state, emit photons, as they contain charged particles, which create electric fields, and (sometimes harder to measure) magnetic fields.
Electromagnetic interactions are mediated by photons, so anything that contains charged particles will emit photons.
Answered by Max Candocia on October 29, 2021
The simple answer is yes. Here is a real life example where this is important.
To produce 'signal', the MRI scanner interacts with protons in the body. Randomly orientated protons become aligned with the powerful magnetic field in the bore of the scanner. A rapidly repeating sequence of radiofrequency pulses – produced by the scanner – then causes 'excitation' and 'resonance' of protons. As each radiofrequency pulse is removed, the protons ‘relax’ to realign with the magnetic field, and as they do so they give off radiofrequency 'signal' which is detected by the scanner and transformed into an image.
Source: https://www.radiologymasterclass.co.uk/tutorials/mri/mri_signal
It may be interesting as well to mention as an additional layer of information that quarks inside the proton also excite similar to an electron. They don't carry the same SPDF class, but instead form [Nucleon] Resonances which are excited states of protons/neutrons. This was the basis behind my PhD work. In this case they will primarily emit particles instead of photons (so pions, Kaons, new nucleons, etc)
Answered by brian h on October 29, 2021
It is good to say that nuclei emit not just photons, but also in some cases, their pretty close relatives - W bosons. That's how beta-decay happen (W boson is quick to decay into electron and neutrino).
Answered by fraxinus on October 29, 2021
Yes, many experiments of the form X(n,n'g)X are performed as a way to measure the energy levels of a nucleus. An incident neutron scatters off the target nucleus and if the incident neutron energy is high enough energy it will leave the target nucleus in an excited state. These states will often decay by emmision of a gamma-ray. The energy level of a nucleus can be determined by measuring the gamma-ray energies.
Answered by Natsfan on October 29, 2021
When nuclei are in an excited state and decay, they often release photons (almost always in the gamma/X-ray energies). A good example of this is the decay of metastable nuclear isomers.
For instance, the decay of Hafnium-178m2 (the 'm' represents a nuclear isomer of Hafnium-178) yields a photon of 2.5 MeV. There are attempts to use this isomer as a weapon. One of the lowest energy nuclear isomers is Thorium-229m, which decays to Thorium-229, releasing a photon with only around 7.6 eV(far ultraviolet!)
Another interesting nuclear isomer is the exceedingly long-lived Tantalum-180m. It is a primordial nuclei (i.e. if you get a sample of Tantalum, it will contain some Tantalum-180m), and has a half-life of over $10^{15}$ years. One decay mode is to drop to Tantalum-180, releasing a 77 keV photon. What is funny about this isomer is that once it drops to the ground state (Tantalum-180), the ground state nucleus is less stable than the excited state isomer, with a 8.1 hour half-life!
Answered by user270049 on October 29, 2021
Yes, excited nuclei emit photons in the form of the highly energetic $gamma$-rays.
That these emissions are much more energetic than the VIS photons emanating from excited electron clouds can be somewhat heuristically understood by looking at the energy spectrum of a particle in a closed $text{1D}$ box, of length $L$:
$$E_n=frac{n^2h^2}{8mL^2}$$
So the emission energies are inversely proportional to system size:
$$Delta E propto frac{1}{L^2}$$
It follows that smaller systems, like nuclei, will emit much more powerful photons than atoms (electron clouds).
The energy content of a VIS photon is about $2text{ }mathrm{eV}$, that of a mid-range $gamma$-photon about $1.2times 10^6text{ }mathrm{eV}$.
Of course neither system above are $text{1D}$ boxes and in reality it is just a useful analogy.
Answered by Gert on October 29, 2021
Nuclei emit gamma rays, which are high energy photons. The photons emitted when electron in an atom changes its energy state are usually in optical spectrum, which are more frequently encountered in technology and real life, which is why they receive more attention in textbooks.
Remarks
To underscore why photons are less "important" for nuclei than for atoms:
Quote
The following quote is from book "Fundamentals in nuclear physics" by Besdevant, Rich and Spiro:
While the numbers (A, Z) or (N, Z) define a nuclear species, they do not
determine uniquely the nuclear quantum state. With few exceptions, a nucleus
(A, Z) possesses a rich spectrum of excited states which can decay to the ground
state of (A, Z) by emitting photons. The emitted photons are often called
gamma-rays. The excitation energies are generally in the MeV range and their
lifetimes are generally in the range of 10^{−9}–10^{−15} s. Because of their
high energies and short lifetimes, the excited states are very rarely seen on Earth
and, when there is no ambiguity, we denote by (A, Z) the ground state of the
corresponding nucleus.
Answered by Roger Vadim on October 29, 2021
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