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**1. Homework Statement**

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

[tex] P = \frac{128}{a_{0}^6} \left[ \int_{0}^{\infty} r^2 e^{-3r/a_{0}} dr\right]^2 [/tex]

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. Homework Equations**

[tex] R_{10} = \frac{2}{\sqrt{a^3}} e^{-r/a} [/tex]

**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

[tex] \psi(x) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) [/tex]

where the [tex]c_{n} = \sqrt{\frac{2}{a}} \int \psi^*(x) \Psi(x,0) dx [/tex]

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

I m also wondering if the Bohr radius would be altered because we have a helium 3 nucleus. In that case

[tex] r_{n} = \frac{n^2 m_{e}}{Z \mu} a_{0} [/tex]

where a0 is the Bohr radius

here n =1, Z = 2, and \mu is still approximately the mass of the elecron. So [itex] a = r_{1} = a_{0}/2 [/itex]??

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

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