Helium Ground State Energy Calculation

This solution outlines the method to determine the ground state energy of a helium atom, utilizing the provided experimental ionization energies for hydrogen and helium.

Key Physics Principles

• The magnitude of the ground state energy for a Hydrogen atom is $13.6$ eV.
• The first ionization energy ($IE_1$) of a Helium atom ($He$) is $24.6$ eV. This is the energy required to remove one electron, forming the Helium ion ($He^+$).
• The $He^+$ ion behaves like a hydrogen atom but with a nuclear charge $Z=2$.

Step-by-Step Energy Calculation

1. Determine the ground state energy of the $He^+$ ion. Using the energy formula for hydrogen-like species, $E_n(Z) = -13.6 \frac{Z^2}{n^2}$ eV:

   For $He^+$ ($Z=2$) in its ground state ($n=1$):

   $E_{He^+, ground} = -13.6 \times \frac{2^2}{1^2} \text{ eV} = -13.6 \times 4 \text{ eV} = -54.4$ eV

   This value, $-54.4$ eV, corresponds to the term $-4(13.6)$ in the options.

2. Use the definition of the first ionization energy ($IE_1$) for Helium:

   $IE_1(He) = E_{He^+, ground} - E_{He, ground}$

3. Rearrange the equation to find the ground state energy of the neutral Helium atom ($E_{He, ground}$):

   $E_{He, ground} = E_{He^+, ground} - IE_1(He)$

4. Substitute the calculated and given values:

   $E_{He, ground} = (-54.4 \text{ eV}) - (24.6 \text{ eV})$

   Which can be expressed using the terms from the options as:

   $E_{He, ground} = (-4 \times 13.6 \text{ eV}) - (24.6 \text{ eV})$

Therefore, the expression $-4(13.6) - 24.6$ accurately represents the ground state energy of the Helium atom in electron volts (eV).