He+ Ionization Energy and Rydberg Equation

Rydberg Equation and Helium Ionization Energy

Introduction

  • The problem involves calculating the ground state ionization energy of He+He^+.
  • He+He^+ is a one-electron system.
  • The Rydberg equation can be used.

Rydberg Equation

  • The relevant equation for a one-electron system is:
    ΔE=−2.18×10−18 J×Z2(1n<em>final2−1n</em>initial2)\Delta E = -2.18 \times 10^{-18} \text{ J} \times Z^2 \left( \frac{1}{n<em>{\text{final}}^2} - \frac{1}{n</em>{\text{initial}}^2} \right)

  • Where:

    • ΔE\Delta E is the change in energy.
    • ZZ is the atomic number (number of protons).
    • nfinaln_{\text{final}} is the final energy level.
    • ninitialn_{\text{initial}} is the initial energy level.

Defining Variables for Helium

  • For helium (He), Z=2Z = 2.
  • Need to determine the values of n<em>initialn<em>{\text{initial}} and n</em>finaln</em>{\text{final}}.

Ground State

  • The term "ground state" refers to the lowest energy state for the electron where ninitial=1n_{\text{initial}} = 1.
  • Analogous to the ground floor being the first floor of a building.

Ionization Energy

  • Ionization means removing the electron.
  • nfinal=∞n_{\text{final}} = \infty (infinity) when the electron is completely removed from the atom.
  • As nn increases, the electron moves further from the nucleus.

Calculation

  • With n<em>initial=1n<em>{\text{initial}} = 1, n</em>final=∞n</em>{\text{final}} = \infty, and Z=2Z = 2, the equation becomes:
    ΔE=−2.18×10−18 J×22(1∞2−112)\Delta E = -2.18 \times 10^{-18} \text{ J} \times 2^2 \left( \frac{1}{\infty^2} - \frac{1}{1^2} \right)

  • Since 1∞2≈0\frac{1}{\infty^2} \approx 0:
    ΔE=−2.18×10−18 J×4×(0−1)\Delta E = -2.18 \times 10^{-18} \text{ J} \times 4 \times (0 - 1)
    ΔE=−2.18×10−18 J×(−4)\Delta E = -2.18 \times 10^{-18} \text{ J} \times (-4)

  • ΔE\Delta E is positive, meaning energy is required to remove the electron.

Result

  • ΔE=8.72×10−18 J\Delta E = 8.72 \times 10^{-18} \text{ J}

  • This is the ionization energy of He+He^+.

Wavelength Calculation

  • The wavelength of light associated with this energy is 23 nanometers.

Electromagnetic Spectrum Context

  • 23 nm wavelength is in the ultraviolet (UV) range of the spectrum.
  • UV light is not within the visible range.

Relevance

  • UV light can lead to skin cancer.
  • Understanding the energy of electrons and their wavelengths has real-world implications.