The Limit of Convergence and Trends in Ionization Energy
- When electrons in an atom absorb energy, they move to higher energy levels.
- These excited states are unstable, so electrons eventually return to lower energy levels, emitting photons with specific energies.
- This produces the line emission spectrum, a series of sharp lines that correspond to the energy differences between levels.
What is Convergence?
- Convergence is similar to what happens in the emission spectrum of hydrogen: as the frequency (or energy) of emitted photons increases, the spectral lines get closer together.
- This phenomenon is called convergence. At the highest frequencies, the lines merge into a continuum, marking the limit of convergence.
- The limit of convergence corresponds to the energy needed to completely remove an electron from the atom, also known as the ionization energy.
Connecting Convergence to Ionization Energy
- The energy of a photon can be calculated using the equation: $$E = h \cdot f$$ where:
- $E$ = energy of the photon (in joules, $\text{J}$)
- $h = 6.63 \times 10^{-34} \text{ J s}$ (Planck's constant)
- $f$ = frequency of the radiation (in hertz, $\text{Hz}$, or $\text{s}^{-1}$)
- The relationship between frequency ($f$) and wavelength ($\lambda$) is given by: $$c = \lambda \cdot f$$ where:
- $c = 3.00 \times 10^8 \ \text{m s}^{-1}$ (speed of light in a vacuum)
- $\lambda$ = wavelength (in meters)
- By combining these equations, you can calculate the ionization energy of an atom if the wavelength at the convergence limit is known: $$E = \frac{h \cdot c}{\lambda}$$
At the convergence limit, the frequency of the emitted photon corresponds to the energy required to ionize the atom.
Example question
Calculating the Ionization Energy of Hydrogen
The spectral lines in the hydrogen emission spectrum converge at a wavelength of $9.12 \times 10^{-8} \ \text{m}$. Calculate the ionization energy of hydrogen in $\mathrm{kJ \cdot mol}^{-1}$.
Solution
- Step 1: Calculate the energy of a single photon. $$E = \frac{6.63 \times 10^{-34} \cdot 3.00 \times 10^8}{9.12 \times 10^{-8}} = 2.18 \times 10^{-18} \, \text{J}$$
- Step 2: Convert to $\mathrm{kJ \cdot mol}^{-1}$ using Avogadro's constant ($N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$). $$ \text{Ionization energy} = 2.18 \times 10^{-18} \cdot 6.02 \times 10^{23} \div 1000 = 1.31 \, \mathrm{kJ \cdot mol}^{-1}$$
- Remember, the limit of convergence directly corresponds to an atom's ionization energy.
- Always convert wavelengths to meters when using the equation $E = \frac{h \cdot c}{\lambda}$.
Trends in Ionization Energy
Definition
Ionization energy
Ionization energy ($IE$) is the minimum energy required to remove an electron from a gaseous atom in its ground state.
Understanding periodic trends in $IE$ helps explain the reactivity of elements.
