Answer:
The energy of these two photons would be the same as long as their frequencies are the same (same color, assuming that the two bulbs emit at only one wavelength.)
Explanation:
The energy 
of a photon is proportional to its frequency 
. The constant of proportionality is Planck's Constant, 
. This proportionality is known as the Planck-Einstein Relation.
.
The color of a beam of visible light depends on the frequency of the light. Assume that the two bulbs in this question each emits light of only one frequency (rather than a mix of light of different frequencies and colors.) Let 
and 
denote the frequency of the light from each bulb.
If the color of the red light from the two bulbs is the same, those two bulbs must emit light at the same frequency: 
.
Thus, by the Planck-Einstein Relation, the energy of a photon from each bulb would also be the same:
.
Note that among these two bulbs, the brighter one appears brighter soley because it emits more photons per unit area in unit time. While the energy of each photon stays the same, the bulb releases more energy by emitting more of these photons.
The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.
<h3><u>Explanation: </u></h3>
A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.
Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.
The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.
Answer:Poopy-di scoop
Scoop-diddy-whoop
Whoop-di-scoop-di-poop
Poop-di-scoopty
Scoopty-whoop
Whoopity-scoop, whoop-poop
Poop-diddy, whoop-scoop
Poop, poop
Scoop-diddy-whoop
Whoop-diddy-scoop
Whoop-diddy-scoop, poop
Explanation:
