0

The power of an electromagnetic wave is proportional to its amplitude squared. (Watts, usually)

But, it is also proportional to its frequency, correct? The higher the frequency, the greater? ( measured in eV).

Is the TOTAL strength of the wave it's watts and electron-volts combined?

So, what is the difference between a laser, say, with a large number of infrared photons versus an x-ray laser with fewer photons, but higher frequency? They have the same total energy, but do they cut the same? Can a low-frequency laser beam, if intense enough, ionized atoms and molecules like individual photons of x-rays do?

Kurt Hikes
  • 4,767
  • 3
  • 19
  • 41

2 Answers2

4

The term: electromagnetic wave belongs to classical electrodynamics, where the energy carried by the wave is described by the Poynting vector.

Electromagnetic waves carry energy as they travel through empty space. There is an energy density associated with both the electric field E and the magnetic field B. The rate of energy transport per unit area is described by the vector

$$\vec{S}=\frac{1}{\mu_{0}}\vec{E}\times\vec{B} $$

There is nothing about frequency in this classical definition of the energy carried by electromagnetic waves.

Frequency enters in the quantum mechanical frame, where it is postulated that the classical wave is built up of photons, particles in the standard model of particle physics, of energy $h\times \nu$ where $h$ is the Planck constant and $\nu$ the frequency of the classical wave emergent from zillions of such photons in quantum mechanical superposition.

The higher the frequency the more energy is carried by the photon, and one can get an average number of photons by dividing the power of the beam by $h\times \nu$. Both in the same units, joules preferably as watts are power, i.e. joules per second, one will get the number of photons per second passing the unit area.

Thomas Abshier
  • 864
anna v
  • 236,935
1

For a classical electromagnetic wave of vector potential amplitude $A$ and frequency $\nu$ the energy is proportional to $\omega^2 A^2$, where $\omega = 2\pi \nu$. For a photon the energy is $\hbar \omega $. The difference is due to normalisation of the wave, which is $A^2 V = \hbar / \omega $, $V$ being the volume. Alternatively, if the wave is stated in terms of $E$, the $E$ field amplitude $\omega A$ absorbs the $\omega$ dependency.

my2cts
  • 27,443