Calculate number of electronic orbitals possible for a Helium atom (Z-2) at Principal
Quantum Number n=3.
b) What is the degenerate energy for all electrons in all different electronic Sommerfeld
orbits.
c) Calculate value of semi-major axis (a) and semi-minor axis (b) for all orbitals.
d) Calculate eccentricity (remember the formula) for all orbits? All orbits are elliptical?

To tackle your question about the electronic structure of a helium atom (Z=2) at the principal quantum number n=3, we can break it down into several parts. Let’s explore each aspect step by step.

1. Number of Electronic Orbitals

The number of electronic orbitals for a given principal quantum number (n) can be determined using the formula:

• Number of orbitals = n²

For helium at n=3:

• n² = 3² = 9

Thus, there are 9 possible electronic orbitals for a helium atom when n=3.

2. Degenerate Energy Levels

Degenerate energy levels refer to the energy states that have the same energy. In the case of the Sommerfeld model, which extends the Bohr model to include elliptical orbits, the energy of an electron in an orbit is primarily determined by the principal quantum number n. For helium, the energy levels can be approximated as:

• E_n = -Z² * 13.6 eV / n²

Substituting Z=2 and n=3:

• E_3 = -2² * 13.6 eV / 3² = -4 * 13.6 eV / 9 = -6.04 eV

Therefore, the degenerate energy for all electrons in the different Sommerfeld orbits at n=3 is approximately -6.04 eV.

3. Semi-Major and Semi-Minor Axes

The semi-major axis (a) and semi-minor axis (b) of an elliptical orbit can be derived from the semi-classical model. For hydrogen-like atoms, the semi-major axis can be calculated using:

• a = n² * a₀ / Z

Where a₀ is the Bohr radius (approximately 0.529 Å). For helium (Z=2) at n=3:

• a = (3² * 0.529 Å) / 2 = (9 * 0.529 Å) / 2 = 2.38 Å

The semi-minor axis can be calculated using the relationship between a and the eccentricity (e) of the orbit:

• b = a * √(1 - e²)

To find b, we first need to calculate the eccentricity.

4. Eccentricity Calculation

The eccentricity (e) of an orbit is given by the formula:

• e = √(1 - (b²/a²))

In the case of circular orbits, e=0, and for elliptical orbits, e is between 0 and 1. For the Sommerfeld orbits, the eccentricity can be approximated based on the quantum number:

• e = 1 - (1/n²)

For n=3:

• e = 1 - (1/3²) = 1 - 1/9 = 8/9 ≈ 0.89

Now, substituting this value back into the equation for b:

• b = 2.38 Å * √(1 - (0.89)²) = 2.38 Å * √(1 - 0.7921) = 2.38 Å * √(0.2079) ≈ 2.38 Å * 0.456 = 1.09 Å

Summary of Results

To summarize:

• Number of electronic orbitals at n=3: 9
• Degenerate energy for electrons in Sommerfeld orbits: -6.04 eV
• Semi-major axis (a): 2.38 Å
• Semi-minor axis (b): 1.09 Å
• Eccentricity (e): 0.89

All orbits are indeed elliptical, as indicated by the eccentricity being greater than 0. This analysis provides a comprehensive understanding of the electronic structure of helium at the specified quantum level.