Particle-Like Nature of Em Radiation
Particle-Like Nature of EM Radiation
When we talk of electromagnetic radiation as a particle, we refer to photons, which are packets of energy. The energy of the photon is related to the wavelength of electromagnetic radiation according to:
Definition: Planck's constant
Planck's constant is a physical constant named after Max Planck.
\[h = \text{6.63} \times \text{10}^{-\text{34}}\text{ J·s}\]The energy of a photon can be calculated using the formula:
\[E = hf\]or
\[E = h\frac{c}{\lambda}\]where \(E\) is the energy of the photon in joules (\(\text{J}\)), \(h\) is Planck's constant, \(c\) is the speed of light, \(f\) is the frequency in hertz (\(\text{Hz}\)) and \(λ\) is the wavelength in metres (\(\text{m}\)).
The higher the frequency of EM radiation, the higher the energy.
Example: Calculating the Energy of a Photon
Question
Calculate the energy of a photon with a frequency of \(\text{3} \times \text{10}^{\text{18}}\) \(\text{Hz}\).
Step 1: Analyze the question
You are asked to determine the energy of a photon given the frequency. The frequency is in standard units and we know the relationship between frequency and energy.
Step 2: Apply the equation for the energy of a photon
\begin{align*} E & = hf \\ & = \text{6.63} \times \text{10}^{-\text{34}}\text{ J·s} \times \text{3} \times \text{10}^{\text{18}}\text{ Hz} \\ & = \text{2} \times \text{10}^{-\text{15}}\text{ J} \end{align*}Step 3: Quote the final result
The energy is \(\text{2} \times \text{10}^{-\text{15}}\) \(\text{J}\)
Example: Calculating the Energy of a Photon II
Question
What is the energy of an ultraviolet photon with a wavelength of \(\text{200}\) \(\text{nm}\)?
Step 1: Analyze the question
You are asked to determine the energy of a photon given the wavelength. The wavelength is in standard units and we know the relationship between frequency and energy. We also know the relationship between wavelength and frequency, the equation for wave speed. The speed of light is a constant that we know.
Step 2: Apply principles
First we determine the frequency in terms of the wavelength.
\begin{align*} c & = f \cdot \lambda \\ f & = \frac{c}{\lambda} \end{align*}We can substitute this into the equation for the energy of a photon, \(E = hf\), allowing us to deduce:
\[E = h\frac{c}{\lambda}\]Step 3: Do the calculation
\begin{align*} E & = h\frac{c}{\lambda} \\ & = \left(\text{6.63} \times \text{10}^{-\text{34}}\text{ J·s}\right)\frac{ \text{3} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$}}{\text{200} \times \text{10}^{-\text{9}}\text{ m}} \\ & = \text{9.939} \times \text{10}^{-\text{10}}\text{ J} \end{align*}Step 4: Quote the final result
The energy of the photon is \(\text{9.39} \times \text{10}^{-\text{10}}\) \(\text{J}\)
This lesson is part of:
Introducing Electromagnetic Waves