Question

Question #85cb9

Answer 1

The electromagnetic spectrum is truly continuous because the energy of every photon is at least slightly uncertain due to the Heisenberg Uncertainty Principle.

The energy of any particular photon is nominally given by

`E=hnu`

where `h` is Planck's constant and `nu` is the frequency of the photon. However, photons (and everything else) are subject the limitations of the Heisenberg Uncertainty Principle, which that states that the product of uncertainties in position and momentum must be greater than or equal to an absolute minimum value:

`(Deltax)(Deltap)>=h/(4pi)`

The momentum of a photon is given by the de Broglie relation

`p=h/lambda`

Using the relationship between wavelength and frequency, `lambda=c/(nu)`, we can easily obtain

`p=h/lambda=(hnu)/c=E/c`
and therefore
`Deltap=(DeltaE)/c`

The Uncertainty relation for a photon then can be expressed as

`(Deltax)(DeltaE)>=(hc)/(4pi)`

This means that any photon that has a finite length (technically, a coherence length where the frequency is well-defined) must have an associated uncertainty in frequency or wavelength. This means that all photons have at least some uncertainty, thereby allowing the electromagnetic spectrum to be truly continuous.

This effect can be observed experimentally by creating extremely short laser pulses, which necessarily have a wide spread in associated frequencies, even if the original light had a relatively narrow spectrum (e.g., from a continuous wave laser).

Answer 2

The em spectrum is truly continuous.

The energy of an emitted photon will depend on the energy difference between two energy levels which can only have specific values i.e. they are quantised.

This energy is given by `E=hf`

h is the Planck Constant and f is the frequency of the emitted radiation.

Whilst energy is quantised for the energy levels in a particular atom there is no restriction on the continuous range of frequencies.