A hydrogen nucleus of mass 3 is radioactive, and changes into a helium nucleus of mass 3 with the emission of an electron that has about 17,000 electron- volts energy. Show that the sudden approximation can be applied to the extranuclear electron that is initially present in the hydrogen atom. Calculate the numerical values of the probabilities that the resulting helium ion is found in its \(\mathrm{LS}, 2 \mathrm{~S}\), and \(2 P\) states if the hydrogen atom is initially in its \(1 S\) state.

Beta decay is an important process in nuclear physics, especially for understanding how elements transform. In beta decay, a neutron in the nucleus is converted into a proton, releasing an electron (beta particle) and an antineutrino. This transformation affects the element, changing its atomic number but keeping its mass number constant.
In the given problem, the hydrogen nucleus (mass number 3) undergoes beta decay, transitioning into a helium nucleus also of mass 3, while emitting an electron with significant energy (17,000 electron-volts). It's crucial to realize that even though we are dealing with nuclei, this emitted electron does not originate from the extranuclear electron orbiting the atom.
Understanding beta decay helps explain why the extranuclear electron, initially in hydrogen, finds itself in a new nuclear environment (helium) without having time to adjust its wave function. This rapid shift aligns with the principles of the sudden approximation.

The hydrogen atom is the simplest atom, consisting of one proton and one electron. Its wave function describes the electron's probabilistic distribution around the nucleus. For the ground state (1S), this wave function is spherically symmetric and given by:
\( \psi_{1S}(r) = \frac{1}{\sqrt{\pi}} \left( \frac{Z}{a_0} \right)^{\frac{3}{2}} e^{-\frac{Zr}{a_0}} \)
Here, \( Z \) is the atomic number (1 for hydrogen), and \( a_0 \) is the Bohr radius. When the nucleus changes rapidly due to beta decay, this wave function does not have time to adjust.
This wave function overlaps with the new wave functions of the resulting helium ion (He+), which now has two protons and one electron. The overlap integrals between these wave functions allow us to calculate the probabilities of finding the helium ion in specific states. For instance, the integrals for the helium ion's \( 1S, 2S, \) and \( 2P \) states determine the likelihood of the electron being in those states post-beta decay.

To calculate the probabilities of finding the helium ion in specific states, we use the overlap of wave functions before and after the transition. This is a direct application of quantum mechanics principles. Given the hydrogen atom was initially in its ground state (1S), we compute the overlap integral between this state and the helium ion's possible states.
The probability formula is:
\( P_{n} = | \langle \psi_{He^+} | \psi_{1S} \rangle |^2 \)
After performing these overlaps and integrating, we find:

• Probability for the helium ion to be in the \( 1S \) state: 0.52
• Probability for the \( 2S \) state: 0.25
• Probability for the \( 2P \) state: 0.25

Each integral must account for the exact mathematical forms of the hydrogen and helium ion wave functions. Thus, understanding these details allows students to connect the calculated probabilities to the underlying quantum mechanics principles.