The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = hv$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of $eV$ for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

(N/A) The energy of a photon is given by the formula:
$E = hv = \frac{hc}{\lambda}$
Where:
$h = \text{Planck's constant} = 6.63 \times 10^{-34} \text{ J s}$
$c = \text{Speed of light} = 3 \times 10^{8} \text{ m/s}$
$\lambda = \text{Wavelength of radiation}$
To convert energy from Joules to $eV$,we divide by $1.6 \times 10^{-19} \text{ J/eV}$.
$E (eV) = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda \times 1.6 \times 10^{-19}} \approx \frac{1.24 \times 10^{-6}}{\lambda} \text{ eV}$.

Part of Spectrum	Wavelength $\lambda$ $(m)$	Energy $E$ (eV)
Radio waves	$10^{3}$	$1.24 \times 10^{-9}$
Microwaves	$10^{-2}$	$1.24 \times 10^{-4}$
Infrared	$10^{-5}$	$0.124$
Visible	$0.5 \times 10^{-6}$	$2.48$
Ultraviolet	$10^{-8}$	$124$
$X$-rays	$10^{-10}$	$1.24 \times 10^{4}$
Gamma rays	$10^{-12}$	$1.24 \times 10^{6}$

The different scales of photon energies are related to the energy level transitions within the source. Higher energy photons (like Gamma rays) correspond to transitions between deep nuclear energy levels,while lower energy photons (like Radio waves) correspond to transitions between atomic or molecular energy levels.