Instantaneous Decay of Tritium into Helium

In summary, the electron in a helium atom would be more likely to be in the ground state if the beta decay caused a proton to be created.
  • #1
uxioq99
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Misplaced Homework Thread
I have been self-studying the MIT 8.04 Introduction to Quantum Mechanics course. This question is not graded, so I have no reservation asking about it on the internet.

Imagine an electron bound by tritium (Z=1). One of the two neutrons undergoes beta decay and becomes a proton, causing the atom to become helium (Z=2). This process happens so quickly that the wave function of the electron is not changed by the process. If the electron is originally in the ground state of tritium, what is the probability that it will be in the ground state of helium after the decay?

My confusion primarily lies with the concept of an electron being "in" an orbital. Although the admissibility of the solution (i.e., energy conservation) leads to a truncation of the Laguerre polynomials and energy discretization, I thought that orbitals were dictated by their wave functions. Then, because the wave functions admit continuous probability distributions, I would think that belonging to an orbital would be more a spectrum than a dichotomy. On the other hand, I recall diagrams of the Laplacian spherical harmonics and their role in producing the wave functions. If I remember correctly, the fact that they were drawn as two dimensional surfaces was more of an artifact of plotting them cleaning.

Before I attempt to answer the question, I would like to crystallize my understanding of the mathematical description of an electron belonging to an orbital.

Thank you all in advance.
 
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  • #2
A helium atom has two electrons. After beta decay of tritium, what is your scenario of getting another electron, or do you think of He3 cation whose Shrodinger equation is easier to solve ?
 
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  • #3
uxioq99 said:
My confusion primarily lies with the concept of an electron being "in" an orbital.
Mathematically, any wave function can be written as a superposition of other wave functions. An orbital is usually meant to be an energy eigenfunction, a state with definite energy. The helium nucleus has twice the charge of the hydrogen nucleus, and the hydrogen-like orbitals of ionized helium are not the same as those of hydrogen. The orbitals have "shrunk", so to speak. So you would have to expand the old wave function in terms of the "scaled down" functions. Or more simply, compute the overlap of lowest H-orbital with the lowest He-orbital and take the squared modulus.
 
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  • #4
uxioq99 said:
I would think that belonging to an orbital would be more a spectrum than a dichotomy.
Not sure what you mean by this, but when they say "ground state" they are telling you you are in the 1S orbital and nowhere ekse,
 
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  • #5
uxioq99 said:
I would think that belonging to an orbital would be more a spectrum than a dichotomy.
First I think the result is a 3He+ ion so the other electron is gone.
The initial state describes the state of the Tritium atom as a whole and it is in an energy eigenstate (the ground state) by fiat (one can suppose whatever one likes). Suddenly the charge of the nucleus doubles because an electron is ejected from it. What is the probability of finding the resulting He+ not in an excited state?
This is an interesting question and the problem as stated is a reasonable approximation, particularly for pedagogical purposes.
 
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  • #6
The electron is described by a state vector ##\lvert \psi \rangle## which belongs to some vector space. Once we choose some basis for this vector space ##\{\lvert \phi_1 \rangle, \lvert \phi_2 \rangle, \dots\}##, we can represent ##\lvert \psi \rangle## as a linear combination of these basis vectors:
$$\lvert \psi \rangle = c_1 \lvert \phi_1 \rangle + c_2 \lvert \phi_2 \rangle + \cdots.$$
One possible basis we could choose consists of the energy eigenstates. Suppose ##\lvert \phi_1 \rangle## corresponds to the ground state, the state with the lowest energy. Then to be in the ground state simply means ##\lvert \psi \rangle## points only along the ##\lvert \phi_1 \rangle## direction. That is, ##\lvert c_1 \rvert = 1## and ##c_2=c_3=\cdots = 0##.
 

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