Exam Notes: Light and Energy

All forms of electromagnetic radiation are light. These include:
- Gamma rays
- Microwaves
- Radio waves
- Visible light
All forms of electromagnetic radiation move at the same speed, i.e., the speed of light ( $c$ ).
The speed of light is approximately $3 \times 10^8$ m/s. This value will be provided on the equation sheet.
Different forms of electromagnetic radiation have different frequencies ( $\nu$ ).
Frequency is the number of cycles per second, measured in hertz (Hz).
Units of hertz:
- $1/s$
- $s^{-1}$
Frequency and energy are directly related. Higher frequency means greater energy.

Wavelength ( $\lambda$ ) is the length of the wave, measured from crest to crest.
Wavelength is typically measured in nanometers (nm) because it's very small.
$1 \times 10^9$ nm = 1 meter
Forms of electromagnetic radiation in order of decreasing wavelength:
1. Radio waves (longest wavelength, do no damage)
2. Microwaves (do no damage)
3. Infrared (does no damage)
4. Visible light
5. UV (Ultraviolet) radiation (mutates cells; watch out for this)
6. X-rays
7. Gamma rays (shortest wavelength)
As wavelength decreases, frequency increases. This is an inverse relationship.
The relationship between wavelength and frequency is defined by the equation: $c = \lambda \nu$ , where $c$ is the speed of light.

Energy ( $E$ ) is measured in joules (J) or kilojoules (kJ).
The equation relating energy and frequency is: $E = h \nu$ , where $h$ is Planck's constant.
$h$ is Planck's constant, and it will be provided on the equation sheet.
As frequency increases, energy increases (direct relationship).

Given a wavelength, you can find the frequency using the equation $c = \lambda \nu$ .
Example: Given a wavelength of 410 nm, find the frequency.
1. Convert nanometers to meters: $410 \text{ nm} \times \frac{1 \text{ m}}{1 \times 10^9 \text{ nm}} = 4.10 \times 10^{-7} \text{ m}$
2. Use the equation $c = \lambda \nu$ to solve for frequency ( $\nu$ ):
  $\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{4.10 \times 10^{-7} \text{ m}} = 7.32 \times 10^{14} \text{ Hz}$
The calculated frequency corresponds to one photon (a unit of light energy).
To find the energy of that light (one photon):
- Use the equation $E = h \nu$ , where $h$ is Planck's constant.
- $E = (6.626 \times 10^{-34} \text{ J s}) \times (7.32 \times 10^{14} \text{ Hz}) = 4.85 \times 10^{-19} \text{ J/photon}$
This is a small amount of energy because it's for only one photon.

To convert from joules per photon to kilojoules per mole:
1. Convert joules to kilojoules: divide by 1000 ( $1 \text{ kJ} = 1000 \text{ J}$ ).
2. Multiply by Avogadro's number ( $6.022 \times 10^{23}$ photons/mole) to convert from per photon to per mole.
- $4.85 \times 10^{-19} \frac{\text{J}}{\text{photon}} \times \frac{1 \text{ kJ}}{1000 \text{ J}} \times \frac{6.022 \times 10^{23} \text{ photons}}{1 \text{ mole}} = \text{kJ/mol}$

1. Aufbau Principle:

Electrons fill the lowest energy orbitals first.
As you move from the nucleus outwards, the energy of orbitals increases.
Electrons will occupy the lowest energy orbitals available before filling higher energy orbitals.

2. Pauli Exclusion Principle:

Each orbital can hold a maximum of two electrons.
These two electrons must have opposite spins (one spin-up, one spin-down).
No two electrons in an atom can have the same set of four quantum numbers (n, l, ml, ms).

3. Hund's Rule:

Electrons will occupy separate orbitals within a subshell (like p or d orbitals) before pairing up in the same orbital.
All singly occupied orbitals within a subshell must have the same spin (e.g., all electrons in a p subshell will have the same spin until the orbitals are filled).
This minimizes electron-electron repulsion, making the configuration more stable.