1. Apr 11, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
Consider the electron in an atom of the heavy isotope of hydrogen,
tritium. The nucleus has charge e, and, apart from a small correction due
to the reduced mass effect, the electron has energies and eigenfunctions that
are identical to those of an ordinary hydrogen atom. However, the nucleus
of the atom, a triton, is unstable and decays by beta-decay to form a
nucleus of 3He. When it does so, the electron in the tritium atom suddenly
finds itself in a new Coulomb potential, the potential due a nucleus with
charge 2e.
Assume that the electron is initially in the ground state of the tritium atom
and show that
$$P = \frac{128}{a_{0}^6} \left[ \int_{0}^{\infty} r^2 e^{-3r/a_{0}} dr\right]^2$$

is the probability that the electron is, after the decay, in the ground state of
the He + ion.
Evaluate the integral and verify that this probability is 0.702.

2. Relevant equations
$$R_{10} = \frac{2}{\sqrt{a^3}} e^{-r/a}$$

3. The attempt at a solution
I was wondering if the radial wavefunctions could eb written as a sum of wavefunctions ... just like inthe case of the square well. That is
$$\psi(x) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x)$$
where the $$c_{n} = \sqrt{\frac{2}{a}} \int \psi^*(x) \Psi(x,0) dx$$

But if that was the case then we can get our probability in pretty much the same way .. by finding $|c_{1})^2$.

I m also wondering if the Bohr radius would be altered because we have a helium 3 nucleus. In that case
$$r_{n} = \frac{n^2 m_{e}}{Z \mu} a_{0}$$
where a0 is the Bohr radius
here n =1, Z = 2, and \mu is still approximately the mass of the elecron. So $a = r_{1} = a_{0}/2$??

The spherical harmonics are normalized so they would not come into the picture here. But would the formula for the c n change ??

Last edited: Apr 11, 2007
2. Apr 11, 2007

### malawi_glenn

Well i would just find the wave function for the ground state of the He+ ion and evaluate the projection of the wave function of the groud state for the tritium (note the isotope shift..).

You should have a table of wave funtions for hydrogentic atoms (i.e. atoms with only one electron) somewere.

3. Apr 11, 2007

### stunner5000pt

yea knowing the waveffunction for the helium+ ion would make things easier but the textbook doesnt give those

it expects we derive it from the ground state of the Hydrogen atom
I know that R10 will change. The new value of a will change to what i get for r1

Is the probability of finding an electron with ground state energy of the the He+ atom found by doing this integral

$$\int_{0}^{\infty} r^2 (R_{10}(r))^2 dr$$
?

4. Apr 11, 2007

### malawi_glenn

Well your last question depends on how your radial wave function looks like, we used two kinds in our course.. one that you must "add" r^2 when you integrate, and one that you doensent.

Frist the hydrogen and helium has the same angular dependence, we are searching in the same state, the 1s.

Second, deriving the radial wave functions for hydrogentic atoms are quite difficult, have your teacher said that you should do so? If not use this:

The radial wave functions i have given, you need to add r^2 when you do the integration to find the electron in a certain state. But scince the question was "the probabilty that the He-ion is in its ground state", we should take the projection of the tritium (H-3) wave function (1s) projected on the He-ion (1s) then integrate. Try :) I'll se if i get the right answer.

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5. Apr 11, 2007

### stunner5000pt

i think you mean $$u_{10} (r) = \frac{R_{10} (r)}{r}$$

i am familiar iwth both

i dont understand why the need for a projection?

6. Apr 11, 2007

### malawi_glenn

We wonder how much of the original wave funtion is in the radial wave function of the helium ion.. ? How much they overlap sort of speaking.

I shall look trough my problem collection, its been a while scince i did quantum mechanics on atom level, i know we had exac this problem too ;)

7. Apr 11, 2007

### malawi_glenn

You also have A.C phillips book?..

8. Apr 11, 2007

### malawi_glenn

I got the right answer, use the wave functions i posted, then SQUARE the result ( iforgot to write that). Then you neglect the reduced mass thing..

The probaility is always the square of the projection. Se pages 72 to 74 in phillps: (4.51) + (4.55)

9. Apr 11, 2007

### stunner5000pt

so when u say projection you mean the Fourier coefficients ... right? the Cn

10. Apr 11, 2007

### stunner5000pt

it also might tak a while for your attachment to be approved
can i find the wavefunctions u posted in phillips

or some other text??

11. Apr 11, 2007

### malawi_glenn

Well, not in this case, you must find out how much of your wave funtion that is covering the state your are searching for. You can think of this as when you are projecting an eigenfunction to your wave function(that perhaps consists of a superposition of many eigenfunctions/states), to find out what the probabilty is to get a certain eigenvalue. This thing is similar.

12. Apr 11, 2007

### malawi_glenn

Mail me malawi_voodoo@hotmail.com i'll mail them to you.

No they are not in the book of phillps, but the problem 9.14 gives you a hint how to get them. you "just add" a Z here and there.. :P

13. Apr 11, 2007

### stunner5000pt

so maybe i should do problem 14 then

14. Apr 11, 2007

### malawi_glenn

Last edited by a moderator: Apr 22, 2017
15. Apr 11, 2007

### stunner5000pt

well i kinda know how those wavefunctions are derived becuase another supplement text (brehm and Mullin) shows that

but i still dont understand why
this question say to look at the problem 8 in chapter 4
i ntaht question we have a particle in an infinite square well of width a/2 and the well suddenly expands to width a. Then we calculate the probability that it will be in the ground state of hte new well by assuming that the wavefunction is a superposition of the original wavefunction (for the a/2 well).

we calculte c1 then by using the formula for the the Fourier coefficients
$$c_{n} = \sqrt{\frac{2}{a}} \int \psi^*(x) \Psi(x,0) dx$$

we need to calculate c1 over there

here we have something similar... the width of the well did change (because the Bohr radius changed). So wecan write the wavefunction as a superpostion of the original wavefunction. But now how do we calculate the cn this time ??

Last edited by a moderator: Apr 22, 2017
16. Apr 11, 2007

### stunner5000pt

it was the Bohr radius that i was not calculating properly

thanks malawi!

17. Apr 11, 2007

### malawi_glenn

You the similar thing in this problem with the atom. But now we know what two states and wave functions that is to be compared..

The well thing is that you take the initial wave function then ask how much of that wave function that is covering the wave function for the ground state for this new well. Then square the result.