Photon Emission and Absorption by Atoms

Quantised Energy Levels

• Electrons in atoms can only exist in specific, discrete energy levels, also known as energy states or orbitals.
• These energy levels are quantised, meaning electrons can only possess certain specific amounts of energy.
• Electrons cannot exist between these energy levels.
• The lowest energy level is called the ground state. Higher energy levels are called excited states.
• An electron can transition between these energy levels by absorbing or emitting energy.

KEY TAKEAWAY: Electrons in atoms can only occupy specific energy levels, not values in between.

Photon Emission

• When an electron transitions from a higher energy level ($E_2$) to a lower energy level ($E_1$), it emits a photon.

• The energy of the emitted photon ($E$) is equal to the difference in energy between the two energy levels:

  $$E = E_2 - E_1$$

• This energy is also related to the photon’s frequency ($f$) and wavelength ($\lambda$) by:

  $$E = hf = \frac{hc}{\lambda}$$

  Where:
  * $h$ is Planck’s constant (\$6.63 \times 10^{-34} \text{ Js}$ or \$4.14 \times 10^{-15} \text{ eVs}$)
  * $c$ is the speed of light (\$3.0 \times 10^8 \text{ m/s}$)

• Emission spectra are produced when excited atoms release photons as their electrons transition to lower energy levels. These spectra consist of discrete bright lines at specific wavelengths, unique to each element.

• The emitted photons have specific energies corresponding to the energy differences between the electron energy levels.

EXAM TIP: Be prepared to calculate the energy, frequency, or wavelength of a photon emitted during an electron transition using $E=hf=hc/\lambda$.

Photon Absorption

• An electron can absorb a photon and transition to a higher energy level only if the photon’s energy exactly matches the energy difference between the two energy levels.
• If the photon energy does not match the energy difference, the photon will not be absorbed.
• Absorption spectra are produced when atoms absorb photons of specific wavelengths from a continuous spectrum. This results in dark lines at those wavelengths, superimposed on the continuous spectrum.
• The absorbed photons have specific energies corresponding to the energy differences between the electron energy levels.

COMMON MISTAKE: Students often confuse emission and absorption. Remember, emission involves an electron losing energy and emitting a photon, while absorption involves an electron gaining energy by absorbing a photon.

The Equation: $\Delta E = hf$

• The fundamental equation that governs photon emission and absorption is:

  $$\Delta E = hf$$

  Where:
  * $\Delta E$ is the change in the electron’s energy (the difference between the initial and final energy levels).
  * $h$ is Planck’s constant.
  * $f$ is the frequency of the emitted or absorbed photon.

• This equation highlights the direct relationship between the energy change of the electron and the frequency of the photon involved.

• Since $f = c/\lambda$, the equation can also be written as:

  $$\Delta E = \frac{hc}{\lambda}$$

• This form relates the energy change to the wavelength of the emitted or absorbed photon.

STUDY HINT: Practice using the equation $\Delta E = hf = hc/\lambda$ with various values of $\Delta E$, $f$, and $\lambda$ to become comfortable with the relationships between these quantities. Remember to use consistent units (Joules or electron-volts for energy, Hertz for frequency, and meters for wavelength).

Electron Standing Waves

• The quantised nature of electron energy levels can be explained by considering electrons as standing waves.
• Only certain wavelengths of electron waves can exist in a stable orbit around the nucleus. These wavelengths must form a standing wave, where the wave interferes constructively with itself.

• The circumference of the electron’s orbit must be an integer multiple of the electron’s wavelength:

  $$2\pi r = n\lambda$$

  Where:
  * $r$ is the radius of the orbit
  * $n$ is an integer (1, 2, 3, …) representing the energy level.
  * $\lambda$ is the de Broglie wavelength of the electron.

• This condition ensures that the electron wave is a standing wave, and the electron can exist in that energy level.

• If the electron wave does not form a standing wave, it will destructively interfere with itself, and the electron cannot exist in that orbit.
• This concept provides evidence for the wave-particle duality of matter, where electrons exhibit both wave and particle properties.

REMEMBER: Electron standing waves explain why only specific energy levels are allowed. If the electron wave doesn’t form a stable standing wave, that energy level is not possible.

Spectra

Emission Spectra

• Consist of bright lines at specific wavelengths.
• Produced when excited atoms emit photons as electrons transition to lower energy levels.
• Unique to each element due to their unique electron energy level structure.
• Used to identify elements in a sample.

Absorption Spectra

• Consist of dark lines at specific wavelengths against a continuous spectrum.
• Produced when atoms absorb photons of specific wavelengths as electrons transition to higher energy levels.
• Unique to each element.
• The dark lines correspond to the same wavelengths as the bright lines in the emission spectrum for that element.

Comparison Table

APPLICATION: Spectroscopy (analyzing spectra) is used in astronomy to determine the composition of stars and galaxies.

Example Calculation

An electron in a hydrogen atom transitions from the $n=3$ energy level to the $n=2$ energy level. The energy difference between these levels is 1.89 eV. Calculate the frequency and wavelength of the emitted photon.

1. Calculate the frequency:

   Using $\Delta E = hf$, we have:
   \$1.89 \text{ eV} = (4.14 \times 10^{-15} \text{ eVs}) \times f$

   $f = \frac{1.89 \text{ eV}}{4.14 \times 10^{-15} \text{ eVs}} = 4.565 \times 10^{14} \text{ Hz}$

2. Calculate the wavelength:

   Using $\Delta E = \frac{hc}{\lambda}$, we have:
   \$1.89 \text{ eV} = \frac{(4.14 \times 10^{-15} \text{ eVs}) \times (3.0 \times 10^8 \text{ m/s})}{\lambda}$

   $\lambda = \frac{(4.14 \times 10^{-15} \text{ eVs}) \times (3.0 \times 10^8 \text{ m/s})}{1.89 \text{ eV}} = 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm}$

VCAA FOCUS: VCAA often includes questions that require you to relate energy level transitions to the wavelengths and frequencies of emitted or absorbed photons. Make sure you can apply the formula $\Delta E = hf = hc/\lambda$ in different problem-solving scenarios.