Quantum Chemistry/Example 6

Using the following equation calculate the ground state energies of H, He+ and Li2+, $$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$, then draw a conclusion on the effect of increasing atomic number on the energy of the ground state.

Solution:

Using the Schrödinger equation (ĤΨ=EΨ) the wave functions of the hydrogen atom have been solved. From the Schrödinger equation, the energy of the quantized states of the hydrogen atom can also be determine, as energy is the eigenvalue of the Schrödinger equation. The equation for energy for each energy level in the hydrogen atom is as follows,

$$

In the equation stated above Z is the nuclear charge, me (9.10938356 x 10-31 kg) is the mass of the electron, e (1.60217662 x 10-19 C) is the elementary charge, ε0 (8.8541878128 x 10-12 Fm-1 ) is the vacuum permittivity, ħ is the reduced Planck's constant and n is the principal quantum number.

The wave functions of the hydrogen atom can be extended to any atom with only one electron. Any atom with only one electron will have the electron in the same energy level as the hydrogen atom as electrons must be placed in the lower energy levels before higher ones. The only thing that will differ between hydrogen and other atoms with one electron is the attraction between the nucleus and the electron. As other atoms such as He+ and Li2+ have a greater number of protons in their nucleus resulting in a greater Coulombic attraction of the electron to the nucleus affecting the orbital size. As a result the energy of the ground state will differ between one electron systems with different nuclear charges.

This can be demonstrated by calculating the ground state energies for H, He+ and Li2+ and seeing how the energies differ. Starting with calculating the ground state of hydrogen where Z=1 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac {(1)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-2.17987225 \times10^{-18} \text {J}$$

$$ 1 \text {eV} = 1.602176634 \times 10^{-19} \text {J}$$

$$ E=-13.60569243 \text {eV}$$

$$1 \text {Hartree}=4.359744722207185 \times 10^{-18} \text {J}$$

$$E=-0.5 \text {Hartree}$$

Then for He+ where Z=2 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac { (2)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-8.71948901 \times10^{-18} \text {J}$$

$$ E=-54.42276978 \text {eV}$$

$$E=-2 \text {Hartree}$$ Finally for Li2+ where Z=3 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac {(3)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-1.96188503 \times10^{-17} \text {J}$$

$$ E=-122.4512322 \text {eV}$$

$$E=-4.5 \text{Hartree}$$

The ground state energy for the hydrogen atom is equal to -0.5 Hartree, for He+ it is -2 Hartree and for Li2+ it is -4.5 Hartree. The energies are increasing in a non-linear fashion which can be attributed to the Z2 term in the numerator. This means as Z increases more energy will be required to ionize a one electron atom.