Electronic Structure of Atoms

To understand the electronic structure of atoms, it's crucial to understand electromagnetic radiation (EMR).
Electromagnetic waves possess characteristic wavelengths ( $\lambda$ ) and frequencies ( $\nu$ ).
Wavelength ( $\lambda$ ) is the distance between corresponding points on adjacent waves.
Frequency ( $\nu$ ) is the number of cycles that pass a point in one second.

All electromagnetic radiation (EMR) moves through a vacuum at the speed of light, denoted as c.
$c = 3.00 \times 10^8 \text{ m/s}$
$c = \lambda \nu$
The SI unit for wavelength ( $\lambda$ ) is the meter (m).
The SI unit for frequency ( $\nu$ ) is s$^{-1}$, which is also known as Hertz (Hz), i.e., $1 \text{ s}^{-1} = 1 \text{ Hz}$ .

Long wavelength ( $\lambda$ ) corresponds to low frequency ( $\nu$ ).
Short wavelength ( $\lambda$ ) corresponds to high frequency ( $\nu$ ).
Wavelength and frequency have an inverse relationship: $\nu \propto \frac{1}{\lambda}$ .

Red light has a wavelength ( $\lambda$ ) of 700 nm. What is its frequency ( $\nu$ )?
1. Convert wavelength to meters:
  $700 \text{ nm} \times \frac{1 \text{ m}}{1 \times 10^9 \text{ nm}} = 7.00 \times 10^{-7} \text{ m}$
2. Use the relationship $\nu = \frac{c}{\lambda}$ :
  $\nu = \frac{3.00 \times 10^8 \text{ m/s}}{7.00 \times 10^{-7} \text{ m}} = 4.29 \times 10^{14} \text{ s}^{-1} = 4.29 \times 10^{14} \text{ Hz}$

All types of electromagnetic radiation travel at the speed of light (c).
$c = \lambda \nu$
$E = h\nu = \frac{hc}{\lambda}$ , where h is Planck’s constant ( $6.626 \times 10^{-34} \text{ J.s}$ ).
Relationships:
- $\nu \propto \frac{1}{\lambda}$
- $E \propto \nu$
- $E \propto \frac{1}{\lambda}$
Longer wavelength ( $\lambda$ ) means lower frequency ( $\nu$ ) and lower energy (E).

High-energy radiations like X-rays have much shorter wavelengths than low-energy radiations like radio waves.
X-rays have high energy, which can cause tissue damage and cancer.

Unit	Symbol	Length (m)	Type of Radiation
Angstrom	Å	$10^{-10}$	X-ray
Nanometer	nm	$10^{-9}$	Ultraviolet, visible
Micrometer	μm	$10^{-6}$	Infrared
Millimeter	mm	$10^{-3}$	Infrared
Centimeter	cm	$10^{-2}$	Microwave
Meter	m	1	TV, radio

Yellow light from a sodium vapor lamp has a wavelength of 589 nm. What is its frequency?
Given: $c = 3.00 \times 10^8 \text{ m/s}$ , $1 \text{ m} = 1 \times 10^9 \text{ nm}$
Solution:
- $\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.09 \times 10^{14} \text{ s}^{-1}$

(a) Laser radiation with $\lambda = 640.0 \text{ nm}$ . Calculate the frequency.
(b) FM radio station broadcasts at $\nu = 103.4 \text{ MHz}$ . Calculate the wavelength.
Answers:
- (a) $4.688 \times 10^{14} \text{ s}^{-1}$
- (b) 2.901 m

A quantum is the smallest amount of energy (photon) that can be emitted or absorbed as electromagnetic radiation.
The relationship between energy and frequency is: $E = h\nu$
where h is Planck’s constant ( $6.626 \times 10^{-34} \text{ J.s}$ ).

The photoelectric effect is the emission of electrons from metal surfaces when light shines on them.
It provides evidence for the particle nature of light and quantization.
Einstein proposed that light travels in energy packets called photons.
The energy of one photon is $E = h\nu$ .
Electrons are ejected only if photons have sufficient energy.

Laser light with $\lambda = 785 \text{ nm}$ . Determine energy and frequency.
Solution:
- Convert nm to m: $785 \text{ nm} = 7.85 \times 10^{-7} \text{ m}$
- Calculate frequency: $\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{7.85 \times 10^{-7} \text{ m}} = 3.82 \times 10^{14} \text{ Hz}$
- Calculate energy: $E = h\nu = (6.63 \times 10^{-34} \text{ J s}) (3.82 \times 10^{14} \text{ s}^{-1}) = 2.53 \times 10^{-19} \text{ J}$

Laser pulse with $\lambda = 532 \text{ nm}$ contains 4.88 mJ of energy. How many photons?
Solution:
- Convert mJ to J: $4.88 \text{ mJ} = 4.88 \times 10^{-3} \text{ J}$
- Calculate energy per photon: $E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J.s})(3 \times 10^8 \text{ m/s})}{532 \times 10^{-9} \text{ m}} = 3.74 \times 10^{-19} \text{ J}$
- Calculate number of photons: \text{#Photons} = \frac{\text{Total energy}}{E_{\text{photon}}} = \frac{4.88 \times 10^{-3} \text{ J}}{3.74 \times 10^{-19} \text{ J/photon}} = 1.31 \times 10^{16} \text{ photons}

Calculate the energy of one photon of yellow light with $\lambda = 589 \text{ nm}$ .
Solution:
- $\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.09 \times 10^{14} \text{ s}^{-1}$
- $E = h\nu = (6.626 \times 10^{-34} \text{ J.s}) (5.09 \times 10^{14} \text{ s}^{-1}) = 3.37 \times 10^{-19} \text{ J}$

(a) Laser emits light with $\nu = 4.69 \times 10^{14} \text{ s}^{-1}$ . What is the energy of one photon?
(b) If the laser emits a pulse of $5.0 \times 10^{17}$ photons, what is the total energy of the pulse?
(c) If the laser emits $1.3 \times 10^{-2} \text{J}$ of energy, how many photons are emitted?
Answers:
- (a) $3.11 \times 10^{-19} \text{ J}$
- (b) 0.16 J
- (c) $4.2 \times 10^{16}$ photons

Quantized: Electrons must gain a specific amount of energy (equal to the energy difference between levels) to move between energy levels.
Energy can be gained from heating (thermal energy) or light (EMR).
Quantum: Smallest amount of energy that can be emitted or absorbed as electromagnetic radiation.
Photon: A quantum of electromagnetic radiation.
Energy of photon: $E = h\nu$ or $E = \frac{hc}{\lambda}$ . $h = 6.626 \times 10^{-34} \text{ J.s}$ .

Rutherford's model: electrons orbit the nucleus like planets orbit the sun. However, this model had issues as charged particles moving in a circle should lose energy, making the atom unstable.
Bohr proposed that electrons are confined to specific energy states called orbits.
Bohr noted the line spectra of certain elements.