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An example of the nucleus of an atom being in an excited state is the Hoyle State, which was a theory devised by the Astronomer Fred Hoyle to help describe the vast quantities of carbon-12 present in the universe, which wouldn't be possible without said Hoyle State.

It's easy to visualise and comprehend the excited states of electrons, because they exist on discrete energy levels that orbit the nucleus, so one can easily see how an electron can excite from one energy level into a higher one, hence making it excited.

However, I can't see how the nucleus can get into an excited state, because clearly, they don't exist on energy levels that they can transfer between, but instead it's just a 'ball' of protons and neutrons.

So how can the nucleus of an atom be excited? What makes it excited?

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Relevant Why-are-pear-shaped-nuclei-possible as you called the nucleus a ball. –  Waqar Ahmad May 17 at 16:47

5 Answers 5

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First you say

It's easy to visualise and comprehend the excited states of electrons, because they exist on discrete energy levels that orbit the nucleus

By way of preparation, I'll note that in introductory course work you never attempt to handle the multi-electron atom in detail. The reason is the complexity of the problem: the inter-electron effects (screening and so on) mean that it is not simple to describe the levels of a non-hydrogen-like atom. The complex spectra of higher Z atoms attest to this.

Later you say

[nuclei] don't exist on energy levels that they can transfer between

but the best models of the nucleus that we have (shell models) do have nucleons occupying discrete orbital states in the combined field of the all the other nucleons (and the mesons that act as the carriers of the "long-range" effective strong force).

This problem is still harder than that of the non-hydrogen-like atoms because there is no heavy, highly-charged nucleus to set the basic landscape on which the players dance, but it is computationally tractable in some cases.

See my answer to "What is an intuitive picture of the motion of nucleons?" for some experimental data exhibiting (in energy space) the shell structure of the protons in the carbon nucleus. In that image you will, however, notice the very large degree of overlap between the s- and p-shell distributions. That is different than what you see in atomic orbitals because the size of the nucleons is comparable to the range of the nuclear strong force.

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It is easy to think of the nucleus as a simple ball because it is so small (about 100,000 times smaller than an atom), but even at this level there is structure. While a different force mediates the interaction among nucleons (the parts of a nucleus), it is analogous to the interactions between nuclei and electrons that gives rise to the structure of atoms. These structures are a consequence of the quantum mechanical rules governing the interactions between particles and fields.

The nucleus of an atom is bound together through the strong nuclear force. This is one of the four fundamental forces, of which electromagnetism is a member. The many states of an given atom are governed by electromagnetic interactions between the electrons and the nucleus of an atom, with a ground state that represents the lowest possible energy configuration of the system, and excited states that are also allowable, but with higher energy values.

Similarly, there are many nuclear states for a given configuration of nuclei. Although mediated through a different fundamental force, there is still a ground state representing the lowest energy configuration of a particular collection of neutrons and protons, and there are many possible excited states as well. These excited nuclear states follow essentially the same rules that excited atomic states, except that the form of the potential term when writing down the Hamiltonian is different.

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Usually, it is after a nucleus has decayed via a $\alpha$- or $\beta$- decay that it is left in a excited state. To see this let us consider the decay of Co-60 into Ni-60.

Co-60 is an unstable nucleus because it has to many neutrons, and therefore it decays in the most probable channel by turning a neutron into proton according to the decay mode

$^{60}$Co$_{27}\rightarrow ^{60}$Ni$_{28} + e^-+\bar{\nu_e}$

or at the nucleon level

$p\rightarrow n+e^-+\bar{\nu_e}.$

The nucleus of Ni$-60$ finds itself in the new configuration of protons and neutrons which does not correspond to the lowest energy level, because the newly created neutron is at a higher energy level due to the decay. So at this stage Ni$-60$ is in an excited state Ni$^*-60$

From this point on the nucleus of Ni$^*_{60}$ will decay to attain its lower energy state. Imagine the similar situation in an atom, where an electron from an inner shell, $E_i$ say, is shot out by a laser beam. The remaining atom is in an excited state because there can be an electron in an outer shell, $E_o$ say, and it will therefore make a transition to the energy level that has just been evacuated.

So Ni$^*-60$ will decay in a cascade with 2$\gamma$ photon emission as shown in the following decays

$^{60}$Ni$^*_{28}\rightarrow$ $^{60}$Ni$^{\prime*}_{28} + \gamma_1$ still an excited state,

$^{60}$Ni$^{\prime*}_{28}\rightarrow$ $^{60}$Ni$_{28} + \gamma_2$ and this is the ground state of Ni-60.

where $E(\gamma_1)$=1.333 MeV and $E(\gamma_2)$=1.173 MeV.

The nucleus of an element can be left in an excited state after an $\alpha$-particle emission as well, and the reasons for the $\gamma$-particle decay are very much the same.

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It is also relatively easy to kick nuclei into excited states with a gentle nudge as in $n\text{(at least 4.5 MeV kinetic energy)} + ^{12}\mathrm{C} \to n + ^{12}\mathrm{C}^*$ followed by the decay of the excited carbon. –  dmckee Mar 3 '13 at 19:42
    
@dmckee Absolutely. Excited nuclei (nuclear isomers) can be manufactured using the mechanism you are describing, as well as using synchrotron irradiation of the nucleus. A very interesting part of applied nuclear physics. –  JKL Mar 3 '13 at 20:52

Point one:

The discrete energy levels for an electron in an atom is due to boundary conditions on Schrodinger equation. Consider a particle in a one dimensional box with potential at boundaries set at infinity, and solve it! Because the potential is infinite outside the box, the wave function must be zero outside the box. The wave function must be continuous, so the wave function at the boundaries must be zero too. From this boundary condition there is a condition on energies which the electron can have.I answered the origin of discrete energy levels. Thus, energy levels do not exist by themselves.

Point two:

When we solve the Schrodinger equation for hydrogen atom, we go to center of mass frame. In this frame our nuclei and electron kinetic operator convert to center of mass kinetic operator and the relative motion kinetic operator. Then, because the mass of nuclei is around 1850 bigger than that of electron the kinetic operator of relative motion reduces to that of electron. Finally, we have the center of mass kinetic operator and relative motion kinetic operator. We assume that the total wave function is the product of wave function of center of mass and relative motion. At the end, we get a discrete energy level for relative motion, which almost is due to motion of electron, and a continuous energy for the center of mass. If we did not use the approximation that the nuclei is much more heavier than electron then those discrete levels would be that of nuclei too. There is another twist to the problem. The nuclei has an internal structure.It is made of quarks. So, those quarks similar to the electron can have their own discrete level of energy. However, in the elementary text books the internal structure of nuclei is not considered.

Summary:

  1. Discrete energy levels are due to boundary conditions on the Schrodinger' s equation.

  2. We separate the kinetic operators of nuclei and electron into center of mass kinetic operator and relative motion operator.

  3. The mass of the nuclei is much more than that of electron. Thus, the relative motion is almost due to the electron.

  4. Now the equation is divided in two parts, center of mass and relative motion and we can assume that the total wave function is the multiplication of the wave function of the two parts.

  5. The solution to the center of mass gives rise to the continuous energy level. The solution to relative motion gives rise to a discrete energy level.

  6. Nuclei has internal structure. They are made of quarks. So, the nuclei can has its own discrete levels similar to the electron.

Perhaps the link below is useful: http://www.pa.msu.edu/~mmoore/Lect28_TwoBodyProb.pdf

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The editor program of this site is horrible. I almost lost my written answer. –  yashar Feb 18 at 13:54
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Many of the basic points you bring up are correct, but they have little bearing on the question (sure quantization comes from boundary conditions, but both of the systems the OP asks about are bound...). Perhaps a bigger issue revolves around your emphasis on the fact that nucleons have their own internal structure: that simply doesn't have much bearing on nuclear structure. At nuclear binding energies, the interaction of nucleons are better described with effective theories like quantum-hadron dynamics and mean field shell models than with fundamental theories like QCD. –  dmckee May 17 at 15:42

While it may be easy to visualize electrons "jumping" up to a higher orbit with another discrete energy level, using that analogy doesn't help grasp the concept of excited nuclei. Instead, think of electrons as oscillators, ones that vibrate around the nucleus in a kind of random way. The region in which they can vibrate around is kind of constrained, and can be understood as the orbital. Each one of these vibrational modes (orbitals) has an energy associated with it, just like a spring vibrating at different frequencies has different energies, or different vibrational modes of speaker have different energies.

Now, imagine that the nucleus is a collection of particles kind of randomly vibrating around. They are pretty well constrained by the forces between them. Suppose that some extra energy is introduced to the nucleus, and causes the particles in the nucleus to vibrate around in a higher energy way, while still staying bound to each other. This is what's called a higher order mode in the vibrational analogy. Just like the electron's wave properties change when it gets or loses energy, or how increasing the frequency of oscillation in a mass-spring system changes it's energy, the extra vibrational energy running around in the nucleus "excites" it into a higher energy state.

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