Quantum of Light - JavaLab

Quantum of Light




  • This simulation assumed that a candle emits all its energy in only one specific wavelength.
  • The power of the candle is assumed to be 50W.
  • The emitted wave diagram is for understanding only and does not represent the actual wavelength and number of photons.

Why aren’t x-rays emitted from a candle?

Energy is needed for light to be emitted.
For example, high-energy x-rays require quite a bit of energy to emit a single photon. Therefore, light with higher energy is more unfavorable to produce than light with lower energy, and is less likely to be emitted.

Every emitted light (photon) can be counted as an integer. That is, light does not exist as a decimal point like 0.5 or 1.3.
At light frequency \(\nu\), the energy of electromagnetic radiation has an integer multiple of \(h\nu\). Where \(h\) is the fundamental constant of nature known as Planck’s constant.(\(h = 6.63\times10^{-34}J\cdot s\))

\[ E = h\nu = \frac{hc}{\lambda } \]

Max Planck (1858-1947) discovered that the energy of light has an integer multiple of \(h\nu\) in 1900. He won the Nobel Prize for this and opened the era of quantum mechanics. Planck’s constant \(h\) is also named after him.
Later, Einstein found out that not only energy, but light(photons) are sparse particles. Einstein also won the Nobel Prize for this achievement.

How many photons are emitted by a ordinary candle?

If the wavelength λ of light is 555 nm (=555 x 10-9m, which is the most sensitive to the eye), the energy of one photon is: \[ \begin{align} E &= \frac{hc}{\lambda } \\ &= \frac{6.626\times 10^{-34} J\cdot s \,\times \, 2.998\times 10^8 m/s}{555\times 10^{-9} m} \\ &\approx 3.58\times 10^{-19} J \end{align} \] Where c is the speed of light.

In general, candles emit most of their energy as heat or infrared(IR) light. And, it emits only a tiny amount of energy as visible light.
The official luminous intensity of one candle is 1 cd (candela). This is a value that releases about \(\frac{4\pi }{683}J\) of energy into space per second.
Dividing this energy by the energy of one photon gives the actual number of photons emitted by the candle in one second.

\[ \frac{\frac{4\pi }{683}J}{3.58\times 10^{-19}J}=5.14\times10^{16} photons \]

* Special thanks to Wolfgang Loeffler for the advice.